A  TEXT  BOOK 


ON 


GRAPHIC  STATICS 


BY 


CHARLES  W.  MALCOLM,  C.  E., 

Assistant  Professor  of  Structural  Engineering,    University  of 

Illinois;  Associate  Member  American  Society  of  Civil 

Engineers;  Member  Society  for  Promotion 

'of  Engineering  Education. 


NEW  YORK  AND  CHICAGO 
THE  MYRON  C.  CLARK  PUBLISHING  CO. 


LONDON 

E.  &  F.  N.  SPON,  LTD.,  57  Haymarket 

1909 


GENERAL 


COPYRIGHT.  1909, 

BY 

CHARLES  W.  MALCOLM, 


PREFACE. 


This  text  was  prepared  for  the  author's  students  in  elemen- 
tary Graphic  Statics  and  Stresses.  It  has  not  been  the  object  of 
the  writer  to  discover  new  principles,  but  rather  to  present  the 
subject  clearly  and  logically.  Many  texts  on  Graphic  Statics  are 
open  to  the  criticism- that  they  have  a  tendency  to  state  principles 
and  give  constructions  without  proofs,  and  the  student  is  there- 
fore compelled  to  memorize  propositions  and  constructions  without 
being  taught  the  underlying  principles.  It  has  been  the  aim  of 
the  writer  to  give  the  proofs,  together  with  full  explanations  of 
the  constructions.  No  attempt  has  been  made  to  give  elaborate 
solutions  which  have  little  or  no  practical  applications.  Particular 
attention  has  been  given  to  the  order  of  presentation;  and  the 
text  has  been  divided  into  chapters,  articles,  and  numbered 
sections  to  facilitate  easy  reference. 

Most  of  the  material  in  Part  I  and  Part  II  has  been  used  in 
printed  form  by  the  author's  students  for  several  years.  It  is 
hoped  that  the  material  in  Part  II,  Part  III,  and  Part  IV  will  be 
found  of  assistance  to  the  practicing  engineer. 

The  author  gratefully  acknowledges  his  indebtedness  to 
Ira  O.  Baker,  Professor  of  Civil  Engineering,  University  of 
Illinois,  for  many  valuable  suggestions  and  criticisms. 

Urbana,  Illinois, 

August,  15,  1909. 


19282S 


CONTENTS. 


PART  I.   GENERAL  PRINCIPLES. 

CHAPTER  I.     DEFINITIONS. 

PAGE 

Dynamics :  Kinetics,  Statics.  Graphic  Statics.  Eigid  Body.  Rest  and 
Motion:  Translation,  Rotation.  Force.  Elements  of  a  Force: 
Magnitude,  Direction,  Line  of  Action,  Point  of  Application.  Con- 
current and  Non-concurrent  Forces.  Coplanar  and  Non-coplanar 
Forces.  Equilibrium.  Equivalence.  Resultant.  Equilibrant.  Com- 
ponents. Composition  of  Forces.  Resolution  of  Forces.  Couple. 
Force  and  Space  Diagrams.  Notation 3 

CHAPTER  II.     CONCURRENT  FORCES. 
ART.  1.     COMPOSITION  OF  CONCURRENT  FORCES 8 

Resultant  of  Two  Concurrent  Forces.     Force  Triangle  and  Force 

Polygon.     Resultant  of  Any  Number  of  Concurrent  Forces. 
ART.  2.     RESOLUTION  OF  CONCURRENT  FORCES „    11 

To  Resolve  a  Force  into  Two  Components. 
ART.  3.     EQUILIBRIUM  OF  CONCURRENT  FORCES 12 

Solution  of  Problems  in  Equilibrium. 

CHAPTER    III.     NON-CONCURRENT    FORCES. 

ART.  1.     COMPOSITION  OF  NON-CONCURRENT  FORCES 16 

Non-concurrent  Forces.  Resultant  of  Two  Non-parallel,  Non-con- 
current Forces:  Forces  Intersecting  Within  the  Limits  of  the 
Drawing;  Forces  Intersecting  Outside  of  the  Limits  of  the  Drawing. 
Resultant  of  Any  Number  of  Non-concurrent  Forces:  Forces 
Having  Intersections  Within  the  Limits  of  the  Drawing;  Forces 
Having  Intersections  Outside  of  the  Limits  of  the  Drawing.  Funic-  . 
ular  Polygon:  Pole,  Pole  Distance,  Rays,  Strings.  Resultant  of 
Any  Number  of  Non-parallel,  Non-concurrent  Forces — Resultant  a 
Couple.  Resultant  of  Any  Number  of  Parallel  Forces.  Closing  of 
the  Funicular  Polygon. 


VI  CONTENTS. 

PAGE 

ART.  2.     KESOLUTION  OF  NON-CONCURRENT  FORCES 23 

Eesolution  of  a  Force  Into  Three  Non-parallel,  Non-concurrent  Com- 
ponents Having  Known  Lines  of  Action.  Resolution  of  a  Force 
Into  Two  Parallel  Components  Having  Known  Lines  of  Action: 
Line  of  Action  of  Known  Forces  Between  the  Lines  of  Action  of 
the  Two  Components;  Line  of  Action  of  Known  Forces  Outside  of 
the  Lines  of  Action  of  the  Two  Components. 

ART.  3.    EQUILIBRIUM  OF  NON-CONCURRENT  FORCES 26 

Conditions  for  Equilibrium  of  Non-concurrent  Forces.  Use  of 
Force  and  Funicular  Polygons  in  Solving  Problems  in  Equilibrium. 
Problem  1,  Parallel  Forces.  Problem  2,  Non-parallel  Forces.  Prob- 
lem 3,  Non-parallel  Forces.  Inaccessible  Points  of  Intersection. 
Problems.  Special  Method.  Problems. 

ABT.  4.    SPECIAL  CONSTRUCTIONS  FOR  FUNICULAR  POLYGONS 34 

Eelation  Between  Different  Funicular  Polygons  for  the  Same 
Forces.  To  Draw  a  Funicular  Polygon  Through  Two  Given  Points. 
To  Draw  a  Funicular  Polygon  Through  Three  Given  Points. 

CHAPTEE  IV.     MOMENTS. 

ART.  1.    MOMENTS  OF  FORCES  AND  OF  COUPLES 38 

Moment  of  a  Force.  Positive  and  Negative  Moments.  Moment 
Area.  Transformation  of  Moment  Area.  Moment  of  the  Eesultant 
of  Two  Concurrent  Forces.  Moment  of  the  Eesultant  of  Any  Num- 
ber of  Concurrent  Forces.  Moment  of  a  Couple.  Moment  of  the 
Eesultant  of  Any  System  of  Forces.  Moment  of  a  System  of 
Forces.  Condition  of  Equilibrium. 

ART.  2.     GRAPHIC  MOMENTS 43 

Graphic  Eepresentation  of  the  Moment  of  a  Force.  Moment  of 
Any  System  of  Forces.  Moment  of  Parallel  Forces.  Problems. 

CHAPTEE  V.  CENTEE  OF  GEAVITY  OF  AEEAS. 

Geometrical  Areas:  Parallelogram,  Triangle,  Quadrilateral,  Circular 
Sector,  Circular  Segment.  Irregular  Areas.  Determination  of  the 
Centroid  of  Parallel  Forces.  Graphic  Determination  of  the  Cen- 
troid  of  a  System  of  Parallel  Forces.  Center  of  Gravity  of  an 
Irregular  Area 47 

CHAPTEE  VI.     MOMENT  OF  INEETIA. 

ART.  1.    MOMENT  OF  INERTIA  OF  PARALLEL  FORCES 52 

Definition.  Moment  of  Inertia  of  a  System  of  Parallel  Forces: 
Culmann's  Method;  Mohr's  Method.  Eelation  Between  Moments 
of  Inertia  About  Parallel  Axes.  Eadius  of  Gyration.  Graphic 
Determination  of  the  Eadius  of  Gyration. 


CONTENTS.  Vll 

PAGE 

ART.  2.    MOMENT  OF  INERTIA  OF  AREAS 58 

Moment  of  Inertia  of  an  Area.  Approximate  Method  for  Finding 
the  Moment  of  Inertia  of  an  Area.  Radius  of  Gyration  of  an  Area. 
Accurate  Method  for  Finding  the  Moment  of  Inertia  of  an  Area. 
Relation  Between  Moments  of  Inertia  of  an  Area  About  Parallel 
Axes.  Moment  of  Inertia  of  an  Area  Determined  from  the  Area  of 
the  Funicular  Polygon  (Mohr's  Method). 


PART  II.  FRAMED  STRUCTURES- 
ROOF  TRUSSES. 

CHAPTER  VII.     DEFINITIONS. 

Framed  Structure.  Types  of  Framed  Structures:  Complete  Framed 
Structure,  Incomplete  Framed  Structure,  Redundant  Framed  Struc- 
ture. Roof  Truss:  Span,  Rise,  Pitch,  Upper  and  Lower  Chords, 
Web  Members,  Pin-connected  and  Riveted  Trusses.  Types  of  Roof 
Trusses:  Fink  Truss,  Quadrangular  Truss,  Howe  Truss,  Pratt 
Truss,  Cantilever  Truss 65 

CHAPTER  VIII.     LOADS. 

ART.  1.    DEAD  LOAD 69 

Construction  of  a  Roof.  Dead  Load:  Roof  Covering;  Purlins, 
Rafters,  and  Bracing;  Roof  Trusses;  Permanent  Loads  Supported 
by  the  Trusses. 

ART.  2.     SNOW  LOAD 72 

ART.  3.     WIND   LOAD 73 

Wind  Pressure.  Wind  Pressure  on  Inclined  Surfaces:  Duchemin's. 
Hutton's,  and  The  Straight  Line  Formulae. 

CHAPTER  IX.     REACTIONS. 

ART.  1.     REACTIONS  FOR  DEAD  AND  SNOW  LOADS 75 

Problem :  Joint  Loads,  Reactions.  Snow  Load  Reactions.  Effective 
Reactions. 

ART.  2.     REACTIONS  FOR  WIND  LOADS 78 

Wind  Load  Reactions:  Truss  Fixed  at  Both  Supports,  Reactions 
Parallel,  Horizontal  Components  of  Reactions  Equal ;  One  End  of 
Truss  Supported  on  Rollers,  Rollers  Under  Leeward  End  of  Truss, 
Rollers  Under  Windward  End  of  Truss. 


Vlll  CONTENTS. 

CHAPTER  X.     STRESSES  IN  ROOF  TRUSSES. 

PAGE 

ART.  1.    DEFINITIONS    AND    GENERAL     METHODS    FOR    DETERMINING 

STRESSES  84 

Definitions:  Tension,  Compression,  Shear.  General  Methods  for 
Determining  Stresses:  Algebraic  Moments,  Graphic  Moments,  Alge- 
braic Resolution,  Graphic  Resolution.  Notation. 

ART.  2.     STRESSES  BY  ALGEBRAIC  MOMENTS 86 

Method  of  Computing  Stresses  by  Algebraic  Moments.     Problem. 

ART.  3.     STRESSES  BY  GRAPHIC  MOMENTS 91 

Method  of  Computing  Stresses  by  Graphic  Moments.     Problem. 

ART.  4.     STRESSES  BY  ALGEBRAIC  RESOLUTION 95 

Method  of  Computing  Stresses  by  Algebraic  Resolution.  Problems: 
Forces  at  a  Joint ;  Forces  on  One  Side  of  a  Section. 

ART.  5.    STRESSES  BY  GRAPHIC  RESOLUTION 99 

Method  of  Computing  Stresses  by  Graphic  Resolution.  Problems: 
Loads  on  Upper  Chord;  Loads  on  Lower  Chord. 

CHAPTER  XI.     WIND  LOAD  STRESSES. 

ART.  1.     BOTH  ENDS  OF  TRUSS  FIXED— REACTIONS  PARALLEL 105 

Problem. 
ART.  2.  BOTH  ENDS  OF  TRUSS  FIXED — HORIZONTAL  COMPONENTS  OF 

REACTIONS  EQUAL 108 

Problem. 
ART.  3.  LEEWARD  END  OF  TRUSS  ON  ROLLERS 110 

Problem. 
ART.  4.  WINDWARD  END  OF  TRUSS  ON  ROLLERS Ill 

Problem. 

CHAPTER  XII.     STRESSES  IN  CANTILEVER  AND  UNSYMMETRI- 
CAL  TRUSSES— MAXIMUM  STRESSES. 

ART.  1.     STRESSES  IN  CANTILEVER  AND  UNSYMMETRICAL  TRUSSES 114 

Stresses  in  a  Cantilever  Truss.  Problem.  Unsymmetrical  Truss — 
Combined  Stress  Diagram.  Problem. 

ART.  2.     MAXIMUM  STRESSES 118 

Problem  1.     Problem  2.     Maximum  and  Minimum  Stresses. 


CHAPTER  XIII.     COUNTERBRACING. 

ART.  1.     DEFINITIONS  AND  NOTATION 125 

Definitions.     Counterbracing.     Notation. 


CONTENTS.  IX 

PAGE 

ART.  2.    STRESSES    IN    TRUSSES    WITH     COUNTERBRACING — SEPARATE 

STRESS  DIAGRAMS 129 

Problem  1,  Truss  with  Parallel   Chords.     Problem  2,   Truss  with 
Non-parallel  Chords. 

ART.  3.     STRESSES     IN     TRUSSES     WITH     COUNTERBRACING — COMBINED  ' 

STRESS  DIAGRAM 139 

Truss  with  Parallel  Chords.     Truss  with  Non-parallel  Chords. 

CHAPTER  XIV.  THREE-HINGED  ARCH. 

Definition.  Reactions  Due  to  a  Single  Load.  Reactions.  Reactions  and 
Stresses  for  Dead  Load.  Wind  Load  Stresses  for  Windward  Seg- 
ment of  Truss.  Wind  Load  Stresses  for  Leeward  Segment  of  Truss.  143 

CHAPTER    XV.     STRESSES     IN     A     TRANSVERSE    BENT     OF    A 

BUILDING. 

Construction  of  a  Transverse  Bent.  Condition  of  Ends  of  Columns: 
Columns  Hinged  at  Base  and  Top;  Columns  Hinged  at  Top  and 
Fixed  at  Base;  Columns  Fixed  at  Top  and  Base.  Dead  and  Snow 
Load  Stresses.  Graphic  Method  for  Determining  Wind  Load  Reac- 
tions. Wind  Load  Stresses — Columns  Hinged  at  Base.  Wind  Load 
Stresses — Columns  Fixed  at  Base 150 

CHAPTER   XVI.      MISCELLANEOUS   PROBLEMS. 

Stresses  in  a  Grand  Stand  Truss.  Stresses  in  a  Trestle  Bent.  Eccen- 
tric Riveted  Connection 160 


PART  III.    BEAMS. 

CHAPTER    XVII.     BENDING    MOMENTS,    SHEARS,    AND    DEFLEC- 
TIONS IN  BEAMS  FOR  FIXED  LOADS. 

ART.  1.     BENDING  MOMENTS  AND  SHEARS  IN  CANTILEVER,  SIMPLE  AND 

OVERHANGING  BEAMS 167 

Definitions.  Bending  Moment  and  Shear  Diagrams  for  a  Cantilever 
Beam:  Cantilever  Beam  with  Concentrated  Loads;  Cantilever 
Beam  with  Uniform  Load.  Bending  Moment  and  Shear  Diagrams 
for  a  Simple  Beam:  Simple  Beam  with  Concentrated  Loads; 
Simple  Beam  with  Uniform  Load.  Bending  Moment  and  Shear 
Diagrams  for  an  Overhanging  Beam :  Overhanging  Beam  with  Con- 
centrated Loads;  Overhanging  Beam  with  Uniform  Load. 


X  CONTENTS. 

PAGE 

ART.  2.     GRAPHIC  METHOD  FOR  DETERMINING  DEFLECTIONS  IN  BEAMS.  .   178 
Explanation    of    Graphic    Method — Constant    Moment    of    Inertia. 
Practical  Application.     Deflection  Diagram — Variable  Moment  of 
Inertia. 

ART.  3.    BENDING  MOMENTS,  SHEARS,  AND  DEFLECTIONS  IN  EESTRAINED 

BEAMS   187 

Definitions.  Bending  Moment,  Shear,  and  Deflection  Diagram  for  a 
Cantilever  Beam.  Bending  Moment,  Shear,  and  Deflection  Diagram 
for  a  Beam  Fixed  at  One  End  and  Supported  at  the  Other.  Sim- 
plified Construction.  Bending  Moment,  Shear,  and  Deflection  Dia- 
grams for  a  Beam  Fixed  at  Both  Ends.  Algebraic  Formulae. 

CHAPTER  XVIII.     MAXIMUM  BENDING  MOMENTS  AND  SHEAES 
IN  BEAMS  FOR  MOVING  LOADS. 

Beam  Loaded  with  a  Uniform  Load:  Maximum  Bending  Moment; 
Maximum  Shear.  Beam  Loaded  with  a  Single  Concentrated  Load: 
Maximum  Bending  Moment;  Maximum  Shear.  Beam  Loaded  with 
Concentrated  Moving  Loads:  Position  for  Maximum  Moment; 
Position  for  Maximum  Shear .  199 


PART  IV.    BRIDGES. 

CHAPTER  XIX.     TYPES  OF  BRIDGE  TRUSSES. 

Through  and  Deck  Bridges.  Types  of  Bridge  Trusses:  Warren,  Howe, 
Pratt,  Baltimore,  Whipple,  Camels-Back,  Parabolic  Bowstring, 
Petit.  Members  of  a  Truss:  Main  Trusses,  Lateral  Bracing, 
Portals,  Knee-braces  and  Sway  Bracing,  Floor  System,  Pedestals, 
Connections  209 

CHAPTER  XX.     LOADS. 

ARI  .  1.    DEAD   LOAD 214 

Weights  of  Highway  Bridges.    Weights  of  Railroad  Bridges. 

ART  2.    LIVE   LOAD 215 

Live  Load  for  Light  Highway  Bridges.  Live  Load  for  Interurban 
Bridges.  Live  Load  for  Railroad  Bridges:  Uniform  Load;  Con- 
centrated Wheel  Loads;  Equivalent  Uniform  Load. 

ART.  3.    WIND  LOAD 219 

Per  Linear  Foot  of  Span.     Per  Square  Foot  of  Surface. 


CONTENTS.  XI 

CHAPTER    XXL     STRESSES     IN     TRUSSES    DUE     TO     UNIFORM 

LOADS. 

PAGE 

ART.  1.     STRESSES  IN  A  WARREN  TRUSS  BY  GRAPHIC  RESOLUTION 221 

Problem.  Dead  Load  Stresses.  Live  Load  Stresses:  Chord  Stresses; 
Web  Stresses;  Maximum  and  Minimum  Live  Load  Web  Stresses. 
Maximum  and  Minimum  Dead  and  Live  Load  Stresses:  Chord 
Stresses;  Web  Stresses.  Loadings  for  Maximum  and  Minimum 
Stresses.  Simplified  Construction  for  Live  Load  Web  Stresses. 

ART.  2.     STRESSES  IN  A  PRATT  TRUSS  BY  GRAPHIC  RESOLUTION. 228 

Problem.  Dead  Load  Stresses.  Live  Load  Stresses:  Chord  Stresses; 
Web  Stresses;  Maximum  and  Minimum  Live  Load  Web  Stresses. 
Maximum  and  Minimum  Dead  and  Live  Load  Stresses:  Chord 
Stresses;  Web  Stresses.  Loadings  for  Maximum  and  Minimum 
Stresses.  Howe  Truss. 

ART.  3.     STRESSES  BY  GRAPHIC  MOMENTS  AND  SHEARS 238 

Problem.  Dead  Load  Chord  Stresses.  Live  Load  Chord  Stresses. 
Dead  Load  Web  Stresses.  Live  Load  Web  Stresses.  Maximum  and 
Minimum  Dead  and  Live  Load  Stresses:  Chord  Stresses;  Web 

Stresses. 

ART.  4.    STRESSES  IN  A  BOWSTRING  TRUSS— TRIANGULAR  WEB  BRACING.  243 
Problem.     Chord  Stresses:     Dead  Load  Chord  Stresses;  Live  Load 
Chord   Stresses.     Web  Stresses:     Dead  Load  Web  Stresses;   Live 
Load  Web  Stresses.     Maximum  and  Minimum  Dead  and  Live  Load 
Stresses:     Chord  Stresses;   Web  Stresses. 

ART.  5.     STRESSES  IN  A  PARABOLIC  BOWSTRING  TRUSS 247 

Problem.  Chord  Stresses:  Dead  Load  Chord  Stresses;  Live  Load 
Chord  Stresses.  Web  Stresses:  Dead  Load  Web  Stresses;  Live 
Load  Web  Stresses.  Maximum  and  Minimum  Dead  and  Live  Load 
Stresses:  Chord  Stresses;  Web  Stresses.  Proof  of  Construction 
Shown  in  Fig.  129:  Stresses  in  Diagonals;  Stresses  in  Verticals. 

ART.  6.     WIND  LOAD  STRESSES  IN  LATERAL  SYSTEMS 255 

Upper  Laterals:  Chord  Stresses;  Web  Stresses.  Lower  Laterals: 
Chord  Stresses,  Fixed  Load,  Moving  Load;  Web  Stresses,  Fixed 
Load,  Moving  Load.  Maximum  and  Minimum  Stresses  in  Upper 
Laterals.  Maximum  and  Minimum  Stresses  in  Lower  Laterals. 

ART.  7.    STRESSES  IN  TRUSSES  WITH  PARALLEL  CHORDS  BY  THE  METHOD 
OF  COEFFICIENTS 259 

Algebraic  Resolution — Method  of  Coefficients.  Loading  for  Maxi- 
mum and  Minimum  Live  Load  Stresses.  Conclusions.  Simplified 
Method.  Coefficients  and  Stresses  in  a  Warren  Truss.  Coefficients 
for  a  Pratt  Truss.  Coefficients  for  a  Baltimore  Truss. 


Xli  CONTENTS. 

CHAPTEE    XXII.     INFLUENCE    DIAGRAMS,    AND    POSITIONS    OF 

ENGINE  AND  TRAIN  LOADS  FOR  MAXIMUM  MOMENTS, 

SHEARS,  AND  STRESSES. 

PAGE 

Influence  Diagrams.  Position  of  Loads  for  a  Maximum  Moment  at 
Any  Point  in  a  Beam,  or  at  Any  Joint  of  the  Loaded  Chord  of  a 
Truss  with  Parallel  or  Inclined  Chords.  Position  of  Loads  for  a 
Maximum  Moment  at  Any  Joint  of  the  Unloaded  Chord  of  a  Truss 
with  Parallel  or  Inclined  Chords.  Position  of  Loads  for  a  Maxi- 
mum Moment  at  a  Panel  Point  of  a  Truss  with  Subordinate 
Bracing.  Position  of  Loads  for  a  Maximum  Shear  at  Any  Point 
in  a  Beam.  Position  of  Loads  for  a  Maximum  Shear  in  Any  Panel 
of  a  Truss  with  Parallel  or  Inclined  Chords.  Position  of  Loads  for 
a  Maximum  Stress  in  Any  Web  Member  of  a  Truss  with  Inclined 
Chords.  Position  of  Loads  for  a  Maximum  Floorbeam  Reaction.  267 


CHAPTER  XXIII.     MAXIMUM  MOMENTS,  SHEARS,  AND  STRESSES 
DUE  TO  ENGINE  AND  TRAIN  LOADS. 

ART.  1.    MAXIMUM  MOMENTS,  SHEARS,  AND  STRESSES  IN  ANY  PARTIC- 
ULAR GIRDER  OR   TRUSS 285 

Maximum  Flange  Stresses  and  Shears  in  a  Plate  Girder:  Flange 
Stresses;  Shears.  Maximum  Chord  and  Web  Stresses  in  a  Pratt 
Truss:  Chord  Stresses;  Web  Stresses. 

ART.  2.     MAXIMUM  MOMENTS,  SHEARS,  AND  STRESSES  IN  GIRDERS  AND 

TRUSSES  OF  VARIOUS  TYPES  AND  SPANS 295 

Load  Line  and  Moment  Diagram.  Application  of  Diagrams  in  Fig. 
149  to  Determining  Maximum  Moments  in  Plate  Girders,  or  at 
Joints  of  the  Loaded  Chord  of  a  Truss  with  Parallel  or  Inclined 
Chords.  Application  of  Diagrams  in  Fig.  149  to  Determining 
Maximum  Moments  at  Panel  Points  in  the  Unloaded  Chord  of  a 
Truss  with  Parallel  or  Inclined  Chords.  Application  of  Diagrams 
in  Fig.  149  to  Determining  Maximum  Shears:  Maximum  Shears 
in  Beams  and  Girders;  Maximum  Shears  in  Trusses.  Application 
of  Diagrams  in  Fig.  149  to  Determining  Maximum  Web  Stresses  in 
Trusses  with  Inclined  Chords.  Determination  of  Maximum  Stresses 
in  a  Truss  with  Subordinate  Bracing — Petit  Tniss. 


GRAPHIC    STATICS. 


INTRODUCTION. 

This  text  will  treat  of  the  general  principles  of  Graphic 
Statics,  and  of  the  application  of  these  principles  to  the  solution 
of  some  of  the  problems  of  especial  interest  to  the  Civil  and 
Structural  Engineer.  For  convenience,  the  subject  will  be  divided 
into  four  parts  as  follows: 

Part  I.  General  Principles. 

Part  II.  Framed  Structures — Roof  Trusses. 

Part  III.  Beams. 

Part  IV.  Bridges. 


Of  THE 

UNIVERSITY 

OF 


PART  I. 

GENERAL  PRINCIPLES. 

CHAPTER  I. 
DEFINITIONS. 

1.  Dynamics  is  the  science  that  treats  of  the  action  of 
forces   upon   a   body   at   rest   or   in    motion.     Its   two   main 
branches  are  kinetics  and  statics. 

Kinetics  treats  of  the  motion  of  bodies  and  of  the  laws  gov- 
erning the  production  of  motion  by  forces. 

Statics  treats  of  the  action  of  forces  under  such  conditions 
that  no  change  of  motion  is  produced  in  the  bodies  acted 
upon.  It  is  therefore  the  science  of  equilibrium — the  science 
by  the  aid  of  which  are  determined  the  forces  necessary  to 
maintain  a  body  in  its  state  of  rest  or  motion,  notwithstand- 
ing disturbing  tendencies. 

2.  Graphic  Statics  has  for  its  object  the  deduction  of  the 
principles  of  statics  and  the  solution  of  statical  problems  by 
means  of  geometrical  constructions. 

3.  Rigid  Body.     A  rigid  body  is  a  body  which  is  incapable 
of  change  in  shape  or  size  when  acted  upon  by  forces.     Such 
a  body  is  only  imaginary,  in  that  all  solids  possess  elasticity 
to  a  greater  or  lesser  degree.     Many  solids,  however,  closely 
approximate  a  condition  of  rigidity  and  for  practical  purposes 
may  be  considered  rigid  if  the  forces  acting  upon  them  are  not 
great  enough  to  cause  rupture.    A  well  designed  and  properly 

3 


4  DEFINITIONS.  Chap.  I. 

constructed  truss  approximates  a  rigid  body,  in  that  it  acts  as 
a  whole  to  resist  external  forces. 

Particle  or  Point.  A  particle  or  point  is  the  smallest  con- 
ceivable rigid  body. 

Every  problem  in  statics  presupposes  the  existence  of  a 
rigid  body  or  particle  upon  which  the  forces  act.  . 

4.  Rest  and  Motion.     Rest  is  the  relation  existing  between 
two  particles  when  a  line  joining  them  does  not  change  either 
in  length  or  in  direction.    Motion  is  the  relation  existing  between 
two  particles  when  the  line  connecting  them  changes  either  in 
length  or  in  direction. 

There  are  two  kinds  of  motion,  viz. :  translation  and 
rotation. 

Translation.  Translation  is  the  motion  corresponding  to  the 
change  in  length  of  the  line  connecting  two  particles.  The 
motion  of  a  body  is  translation  when  every  point  in  the  body 
travels  in  a  straight  line. 

Rotation.  Rotation  is  the  motion  of  a  body  when  all  points 
in  the  body,  that  change  their  positions  at  all,  describe  con- 
centric circles  in  parallel  planes.  The  common  normal  to 
these  planes,  which  contains  the  centers  of  all  the  circles,  is 
called  the  axis  of  rotation.  If  the  axis  of  rotation  is  fixed  in 
position,  the  motion  is  pure  rotation. 

There  may  be  compound  translation,  as  that  of  a  body 
sliding  across  a  moving  car;  or  compound  rotation,  as  that  of 
the  earth  revolving  around  its  own  axis  and  also  about  the 
sun ;  or  combined  translation  and  rotation,  as  that  of  a  ball 
rolling  along  a  straight  path. 

Two  bodies  may  also  b'e  at  rest  with  respect  to  each  other 
and  in  motion  with  respect  to  a  third  body.  Rest,  then,  means 
motionlcssness  with  respect  to  a  definite  body  of  reference, 
which  in  this  work  is  understood  to  be  the  earth. 

5.  Force.     Force  is  an  action  exerted  .upon  a  body  tending 
to  chanee  its  state  of  rest  or  motion. 

6.  Elements  of  a  Force.     Tn   order  that  a  force  may  be 
completely  known,  four  characteristics  of  it,  called  elements, 
are  essential,  viz. : 


DEFINITIONS.  5 

1 .  Magnitude. 

2.  Direction. 

3.  Line  of  Action. 

4.  Point  of  Application. 

7.  Magnitude.     The   magnitude   of   a   force   is   given    by 
stating,  numerically,  the  ratio  of  its  effectiveness  in  producing 
motion  to  that  of  the  unit  force.     The  magnitude  of  a  force 
may  be  expressed,  graphically,  by  the  length  of  a  line,  the 
magnitude  of  the  force  being  the  ratio  of  the  given  length  to 
the  unit  length. 

8.  Direction.     Direction  is  the  specification  as  to  which 
of  the  two  ways  along  the  line  the  force  tends  to  produce 
motion.      Direction    is    expressed,    graphically,   by   an    arrow 
placed  on   the   line  representing  the   force,   indicating  which 
way  the  force  tends  to  produce  motion. 

9.  Line  of  Action.     The  line  of  action  of  a  force  is  the 
path  along  which  the   force  tends  to  produce  motion.     The 
line  of  action  is  expressed,  graphically,  by  the  position  of  the 
line  representing  the  force. 

10.  Point  of  Application.     The  point  of  application  is  the 
place   (considered  as  a  point)   where  the  force  is  brought  to 
bear  upon  the  body.     The  point  of  application  is  on  the  line 
of  action  of  the  force,  and  together  with  its  position  locates 
the  line. 

11.  Concurrent  and  Non-concurrent  Forces.      Concurrent 
forces  are  those  whose  lines  of  action  meet  in  a  point.   Non-con- 
current forces  are  those  whose  lines  of  action  do  not  meet  in  a 
point.    Whether  or  not  a  given  system  of  forces  is  concurrent 
or  non-concurrent  affects  rotation   only,   since   translation   is 
entirely  independent  of  the  position  of  the  point  of  application 
of  the  forces. 

12.  Coplanar  and  Non-coplanar  Forces.    -Co planar   forces 
are  those  whose  lines  of  action  lie  in  the  same  plane.  Non-coplanar 
forces  are  those  whose  lines  of  action  do  not  lie  in  the  same 
plane.     Except  where  statements  are  of  a  general  character, 
coplanar  forces  only  will  be  treated  in  this  work. 

13.  Equilibrium.     A  system  of  two  or  more  forces  is  in 


6  DEFINITIONS.  Chap.  I. 

equilibrium  when  the  combined  effect  of  the  forces  produces 
no  change  in  the  body  with  respect  to  its  state  of  rest  or 
motion. 

14.  Equivalence.     Two   forces   or  systems  of  forces  are 
equivalent  when  they  have  identical  effects  upon  the  body 
acted  upon,  with  respect  to  its  state  of  rest  or  motion. 

15.  Resultant.     The  resultant  of  a  system  of  forces  is  the 
simplest   system   which   is   equivalent   to   the   given   system. 
Usually   the   given   system    is   equivalent   to  a   single   force, 
although  this  is  not  always  the  case. 

16.  Equilibrant.     A  single  force,  or  the  simplest  system, 
which  will  exactly  neutralize  the  effect  of  a  system  of  forces, 
is  called  the  equilibrant  of  the  system.     The  equilibrant  is 
numerically  equal  to  the  resultant,  but  acts  in  an  opposite 
direction. 

17.  Components.     Any  one  of  a  system  of  forces  having 
a  given  force  for  its  resultant  is  called  a  component  of  that 
force.     It  is  evident  that  a  force  may  have  any  number  of 
components. 

18.  Composition  of  Forces.     Composition  of  forces  is  the 
process  of  finding,  for  a  given  system  of  forces,  an  equivalent 
system  having  a  smaller  number  of  forces  than   the  given 
system.     The  process  of  finding  a  single  force  to  replace  the 
given  system  is  the  most  important  case  of  composition. 

19.  Resolution  of  Forces.     Resolution  of  forces  is  the  pro- 
cess of  finding,  for  a  given  system  of  forces,  an  equivalent 
system  having  a   greater  number  of   forces   than   the   given 
system.    The  process  of  finding  two  or  more  forces  which  are 
equivalent  to  a  given  force  is  the  most  important  case  of  reso- 
lution. 

20.  Couple.      A    couple    consists    of    two    equal,    parallel 
forces,  opposite  in  direction,  having  different  lines  of  action. 
The  arm  of  the  couple  is  the  perpendicular  distance  between 
the  lines  of  action  of  the  two  forces. 

21.  Force  and  Space  Diagrams.     In  the  solution  of  prob- 
lems in  graphic  statics,  it  is  usually  most  convenient  to  draw 
two  separate  figures,  one  of  which  shows  the  forces  in  magni- 


OF  THE 

UNIVERSITY 

OF 


tude  and  direction  and  the  other  in  line  of  action.  The  former 
is  called  the  force  diagram,  and  the  latter  the  space  diagram. 

22.  Notation.  In  solving  problems,  graphically,  it  is 
often  convenient  to  drawr  both  the  force  and  the  space  dia- 
grams, and  these  diagrams  are  so  related  that  for  every  line 
in  one  there  is  a  corresponding  line  in  the  other.  The  solution 
is  greatly  facilitated  if  a  convenient  system  of  notation  is 
adopted  for  the  diagrams. 

In  the  force  diagram,  each  line  represents  a  force  in  mag- 
nitude and  direction,  and  this  line  will  be  designated  by  plac- 
ing a  capital  letter  at  each  e: 
tremity  of  the  line.  In  the 
space  diagram,  the  correspond- 
ing line  represents  the  line  of 
action  of  the  force,  and  this 
line  of  action  will  be  marked 
by  the  corresponding  small  B 

letters,  one  on  each  side  of  the  FlG-  1- 

line.  An  arrow  placed  on  the  line  indicates  the  direction  of 
the  force.  The  sequence  of  the  letters  representing  the  force 
also  indicates  the  direction  of  the  force ;  thus,  the  force  repre- 
sented by  AB  acts  in  a  direction  from  A  towards  B.  This 
system  of  notation  is  illustrated  in  Fig.  I,  AB  representing  the 
force  in  magnitude  and  direction,  while  its  line  of  action  is 
marked  by  the  letters  a  b  placed  as  shown. 


CHAPTER  II. 

CONCUKEENT   FOKCES. 

The  subject  matter  given  in  this  chapter  will  be  divided 
into  three  articles,  as  follows:  Art.  i,  Composition  of  Concur- 
rent Forces ;  Art.  2,  Resolution  of  Concurrent  Forces ;  Art.  3, 
Equilibrium  of  Concurrent  Forces. 

ART.  I.    COMPOSITION  OF  CONCURRENT  FORCES. 

23.  Resultant  of  Two  Concurrent  Forces.  Given  the  two 
concurrent  forces  represented  in  magnitude  and  direction  by 
the  lines  BA  and  BC.  It  is  required  to  find  their  resultant. 


fort 


FIG.    2. 


Let  BA  and  BC  (Fig.  2,  a)  represent  two  concurrent  forces 
in  magnitude  and  direction,  acting  at  B.  It  is  required  to  find 
their  resultant  BD.  Complete  the  parallelogram  of  forces 
(Fig.  2,  a)  by  drawing  the  line  CD  parallel  to  BA,  and  AD 
parallel  to  BC;  then  the  line  BD,  connecting  the  points  B  and 
D,  will  represent  the  resultant  of  the  two  forces  in  magnitude. 
The  direction  of  the  resultant  will  be  as  indicated  by  the  arrow 
placed  on  the  line  BD.  (The  proof  of  the  above  will  not  be 

8 


Art.  1.  COMPOSITION    OF    CONCURRENT    FORCES.  9 

given  here  as  it  may  be  found  in  any  elementary  treatise  on 
mechanics.) 

It  is  unnecessary  to  construct  the  entire  force  parallelo- 
gram, for,  draw  AB  (Fig.  2,  b)  equal  and  parallel  to  BA  (Fig. 
2,  a),  and  BC  equal  and  parallel  to  BC.  Then  AC  will  be  equal 
and  parallel  to  BD;  for,  by  construction,  the  triangle  ABC  is 
equal  to  the  triangle  ABD  and  has  its  corresponding  sides 
parallel ;  hence,  AC  is  equal  and  parallel  to  BD  the  required 
resultant.  Likewise,  CA  (Fig.  2,  c)  is  the  resultant  of  the  two 
concurrent  forces  BA  and  BC.  It  should  be  noted  that  the  two 
forces  act  around  the  triangle  in  the  same  direction,  and  that 
the  resultant  acts  in  a  direction  opposite  to  them. 

It  is  thus  seen  that  any  two  concurrent  forces  may  be  com- 
bined into  a  single  resultant  force,  and  further  that  it  is  imma- 
terial in  which  order  the  forces  are  taken  so  long  as  they  act 
in  the  same  direction  around  the  triangle. 

24.  Force    Triangle    and    Force    Polygon.     The    triangle 
shown  in  either  Fig.  2,  b,  or  Fig.  2,  c,  is  called  a  force  triangle, 
and  its  principles  form  the  basis  of  the  science  of  graphic  statics. 
If  the  figure  has  more  than  three  sides  it  is  called  a  force  polygon. 

25.  Resultant  of  Any  Number  of  Concurrent  Forces.    The 
resultant  of  any  number  of  concurrent  forces  may  be  found 
(i)  by  the  method  of  the  force  triangle,  or  (2)  by  the  method 
of  the  force  polygon. 

( i )     Solution  by  Force  Triangle.     Let  AB,  AC,  AD,  AE,  and 
AF  (Fig.  3)  represent  in  magnitude  and 
direction  a  system  of  forces  meeting  at  A. 
It  is  required  to  find  their  resultant  R. 

The  method  explained  in  §  23  may 
be  applied  to  any  number  of  forces. 
Commencing  at  B  (Fig.  3),  the  extrem- 
ity of  the  force  AB,  draw  the  line  BG 
equal  and  parallel  to  the  force  AC ; 
then  AG,  acting  in  the  direction 
shown,  is  the  resultant  of  the  forces 
AB  and  AC.  In  like  manner,  from  the 
point  G,  draw  the  line  GH  equal  and  FlG  3 


10 


CONCURRENT  FORCES. 


Chap.  II. 


parallel  to  the  force  AD ;  then  AH,  acting  in  the  direction  shown, 
is  the  resultant  of  the  forces  AG  and  AD,  or,  since  AG  is  the 
resultant  of  AB  and  AC,  then  AH  is  also  the  resultant  of  the 
forces  AB,  AC,  and  AD.  In  like  manner,  AI  is  the  resultant  of 
the  forces  AB,  AC,  AD,  and  AE ;  and  AJ,  acting  in  the  direction 
shown,  is  the  required  resultant  R  of  the  given  system  of  forces 
AB,  AC,  AD,  AE,  and  AF. 

(2)  Solution  by  Force  Polygon.  Let  AB,  BC,  CD,  DE,  and 
EF  (Fig.  4)  represent  in  magnitude  and  direction  a  system  of 
forces  meeting  at  O.  It  is  required  to  find  their  resultant  R. 


FIG.  4. 

The  given  system  of  forces  is  represented  in  magnitude 
and  direction  in  Fig.  4,  b,  and  in  line  of  action  in  Fig.  4,  a. 
Commencing  at  any  point  A  (Fig.  4,  b),  draw  in  succession  the 
lines  AB,  BC,  CD,  DE,  and  EF,  parallel  respectively  to  the 
lines  of  action  ab,  be,  cd,  de,  and  ef,  representing  the  given 
forces  in  magnitude  and  direction,  each  force  beginning  at  the 
end  of  the  preceding  one.  Then  AF,  the  line  connecting  the 
starting  point  of  the  force  polygon  with  the  end  of  the  last 
force,  represents  in  magnitude  and  direction  the  required 
resultant  R.  For,  inserting  the  dotted  lines  AC,  AD,  and  AE, 
which  divide  the  polygon  into  triangles,  it  is  evident  that  AC, 
acting  in  the  direction  shown,  represents  in  magnitude  and 
direction  the  resultant  of  the  two  forces  AB  and  BC.  Like- 
wise, AD  represents  in  magnitude  and  direction  the  resultant 
of  the  forces  AC  and  CD,  or  in  other  words,  AD  is  the  result- 


Art.  2.  RESOLUTION   OF   CONCURRENT   FORCES.  11 

ant  of  the  forces  AB,  BC,  and  CD.  In  like  manner,  AE  is  the 
resultant  of  the  forces  AB,  BC,  CD,  and  DE;  and  AF,  acting 
in  the  direction  shown,  represents  in  magnitude  and  direction 
the  required  resultant  R  of  the  given  system  of  forces.  The 
line  of  action  of  R  passes  through  O,  and  its  direction  is  as 
indicated  by  the  arrow. 

It  will  be  seen  by  referring  to  the  force  polygon  that  all 
the  forces  except  the  resultant  act  around  the  force  polygon 
in  the  same  direction,  and  that  the  resultant  acts  in  a  direction 
opposite  to  them.  By  taking  the  forces  in  a  different  order 
from  that  shown  in  Fig.  4,  it  will  be  found  that  the  order  in 
which  the  forces  are  taken  is  immaterial,  so  long  as  the  given 
forces  act  around  the  force  polygon  in  the  same  direction; 
since  the  positions  of  the  initial  and  final  points  remain  the 
same. 

If  the  points  A  and  F  coincide,  the  force  polygon  is  said  to 
be  closed. 


ART.  2.     RESOLUTION  OF  CONCURRENT  FORCES. 

26.  It  is  readily  seen  from  what  has  been  given  in  the 
preceding  sections  that,  to  resolve  a  given  force  into  any  num- 
ber  of   components,   it   is   only   necessary   to   draw   a   closed 
polygon,  one  side  of  which  represents  in  magnitude  the  given 
force  and  is  parallel  to  its  line  of  action ;  then  the  other  sides 
of  the  polygon  will  be  parallel  to,  and  will  represent  in  mag- 
nitude, the  components  into  which  the  given  force  is  resolved. 
The  given  force  will  act  around  the  force  polygon  in  a  direc- 
tion opposite  to  that  of  the  components. 

27.  To  Resolve  a  Given  Force  Into  Two  Components.     It 
is  evident  that  this  problem  is  indeterminate  unless  other  con- 
ditions are  imposed;  as  an  infinite  number  of  triangles  may 
be  drawn  with  one  side  of  the  given  length. 

There  are  four  cases  of  the  resolution  of  a  force  into  two 
components,  corresponding  to  the  four  cases  of  the  solution  of 
a  plane  triangle,  which  may  be  stated  as  follows: 


12 


CONCURRENT   FORCES. 


Chap.  II. 


(1)  It  is  required  to  resolve  a  given  force  into  two  com- 
ponents which  are  known  in  line  of  action  only. 

(2)  It  is  required  to  resolve  a  given  force  into  two  com- 
ponents which  are  known  in  magnitude  only. 

(3)  It  is  required  to  resolve  a  given  force  into  two  com- 
ponents, one  of  which  is  known  in  line  of  action  only,  and  the 
other  in  magnitude  only  (two  solutions). 

(4)  It  is  required  to  resolve  a  given  force  into  two  com- 
ponents, one  of  which  is  completely  known,  while  the  other  is 
completely  unknown. 

The  solution  of  the  first  case  is  given  below,  the  other 
three  cases  being  left  for  the  student  to  solve. 

Case  I.  Let  AB  (Fig.  5)  represent  in  magnitude  and  direc- 
tion the  known  force,  and  let  ac  and  cb  represent  the  lines  of 
action  of  the  two  components  meeting  at  O.  It  is  required  to 
find  the  unknown  elements  of  the  two  components. 


B 


PIG    5. 


From  the  extremities  of  the  known  force  AB  (Fig.  5),  draw 
the  lines  AC  and  CB,  parallel  respectively  to  ac  and  cb,  inter- 
secting at  the  point  C ;  then  AC  and  CB  represent  in  magni- 
tude the  two  components.  The  directions  of  these  components 
are  shown  by  the  arrows. 


ART.  3.     EQUILIBRIUM  OF  CONCURRENT  FORCES. 

28.  Equilibrium  of  Concurrent  Forces.  A  system  of  con- 
current forces  is  in  equilibrium  if  the  resultant  of  the  system 
is  equal  to  zero ;  since  this  is  the  condition  which  must  exist  if 
no  motion  takes  place.  Referring  to  Fig.  4,  it  is  seen  that  if 


Art.  3.  EQUILIBRIUM  OJF  CONCURRENT  FORCES.  13 

the  resultant  is  zero,  the  force  polygon  is  closed ;  hence  the 
following  proposition:  If  any  system  of  concurrent  forces  is 
in  equilibrium,  the  force  polygon  must  close;  and  conversely,' 
if  the  force  polygon  closes  and  the  forces  act  in  the  same 
direction  around  the  polygon,  the  system  is  in  equilibrium. 
This  is  equivalent  to  the  algebraic  statements  that  the  summa- 
tion of  the  horizontal  components  of  the  forces  is  equal  to 
zero,  and  that  the  summation  of  the  vertical  components  is 
equal  to  zero. 

If  the  system  of  forces  is  not  in  equilibrium,  the  resultant 
is  represented  in  magnitude  and  direction  by  the  closing  line 
of  the  force  polygon,  this  resultant  acting  around  the  force 
polygon  in  a  direction  opposite  to  the  other  forces.  If  a  force 
having  the  same  magnitude  but  acting  in  an  opposite  direction 
is  substituted  for  the  resultant,  the  system  is  then  in  equili- 
brium. It  has  been  shown  that  the  resultant  acts  around  the 
force  polygon  in  a  direction  opposite  to  the  given  forces ;  and 
therefore  the  force  that  will  hold  the  given  system  in  equi- 
librium, called  the  equilibrant,  acts  around  the  force  polygon 
in  the  same  direction  as  the  given  forces. 

29.  Solution  of  Problems  in  Equilibrium.  The  fact  that 
the  force  polygon  for  a  system  of  concurrent  forces  is  closed  if 
that  system  is  in  equilibrium,  furnishes  a  method  for  solving 
problems  in  equilibrium  when  all  the  elements  of  the  forces 
are  not  known. 

The  following  problems  illustrate  the  general  principles 
employed  in  the  solution  of  concurrent  forces  in  equilibrium : 

Problem  I.  Given  a  system  of  five  concurrent  forces  in  equi- 
librium, three  of  which  are  completely  known ;  of  the  remain- 
ing two,  one  is  known  in  magnitude  only,  and  the  other  in  line 
of  action  only.  It  is  required  to  find  the  unknown  elements  of 
the  two  forces.  Give  two  solutions. 

Problem  2.  Given  a  system  of  five  concurrent  forces  in  equi- 
librium, three  of  which  are  completely  known,  the  other  two 
being  known  in  line  of  action  only.  It  is  required  to  find  the 
unknown  elements  of  the  two  forces. 

Problem  j.     Given  a  system  of  five  concurrent  forces  in  equi- 


14 


CONCURRENT  FORCES. 


Chap.  II. 


librium,  one  of  which  is  completely  unknown.  It  is  required 
to  fully  determine  the  unknown  force. 

Problem  4.  Given  a  system  of  five  concurrent  forces  in  equi- 
librium, three  of  which  are  completely  known,  the  other  two 
being  known  in  magnitude  only.  It  is  required  to  find  the 
unknown  elements  of  the  two  forces.  Give  two  solutions. 

The  solution  of  the  first  problem  is  given  below,  the  others 
being  left  for  the  students  to  solve. 

Problem  I.  Let  the  forces  AB,  BC,  and  CD  (Fig.  6)  be  com- 
pletely known,  let  DE  be  known  in  magnitude  only,  and  EA 
in  line  of  action  only.  It  is  required  to  find  the  unknown  ele- 
ments of  the  two  forces. 


FIO.  6. 

First  construct  the  portion  of  the  force  polygon  ABCD 
(Fig.  6),  using  as  sides  the  known  forces  AB,  BC,  and  CD, 
taking  care  that  the  forces  act  progressively  around  the  force 
polygon.  Complete  the  polygon  by  drawing  the  line  EA 
through  A,  parallel  to  the  known  line  of  action  of  EA.  Then, 
using  D  as  a  center  and  the  known  magnitude  of  DE  as  a 
radius,  draw  an  arc  of  a  circle  intersecting  EA  at  the  point  E. 
The  lines  DE  and  EA  will  then  represent  in  magnitude  the 
forces  DE  and  EA,  these  forces  acting  around  the  force 
polygon  in  the  same  direction  as  the  known  forces  AB,  BC, 
and  CD.  The  entire  force  polygon  is  ABCDE,  a  closed 
polygon ;  and  since  all  the  forces  act  around  the  force  polygon 
in  the  same  direction,  they  are  in  equilibrium.  ABCDE'  is 
also  a  true  form  of  the  force  polygon,  and  gives  correct  values 


Art.  3.  EQUILIBRIUM  OF  CONCURRENT  FORCES.  15 

for  the  unknown  forces ;  since  all  the  conditions  of  the  problem 
are  fulfilled,  i.e.,  the  force  polygon  is  closed,  and  the  forces 
act  continuously  around  the  polygon.  Another  solution,  which 
gives  different  values  for  the  unknown  forces,  is  indicated  by 
the  force  polygon  ABCDEl. 


CHAPTER  III. 

NON-CONCURRENT  FORCES. 

The  subject  matter  given  in  this  chapter  will  be  divided 
into  four  articles,  as  follows:  Art.  i,  Composition  of  Non- 
concurrent  Forces ;  Art.  2,  Resolution  of  Non-concurrent 
Forces ;  Art.  3,  Equilibrium  of  Non-concurrent  Forces ;  and 
Art.  4,  Special  Constructions  for  Funicular  Polygons. 

ART.  i.     COMPOSITION  OF  NON-CONCURRENT  FORCES. 

30.  Non- concurrent  Forces.     In   many  engineering  prob- 
lems, the  forces  acting  upon  trie  body  or  structure  do  not  meet 
at  a  point,  but  are  applied  at  different  points  along  the  struc- 
ture.   Such  forces  are  non-concurrent,  and  their  lines  of  action 
may,  or  may  not,  be  parallel. 

31.  Resultant     of     Two     Non-parallel,     Non-concurrent 
Forces.       The   determination   of   the   resultant   of   two   non- 
parallel,  non-concurrent  forces  requires  one  of  two  somewhat 
different   methods  of  solution,   depending  upon   whether   the 
given  forces  intersect  inside  or  outside  the  limits  of  the  drawing. 

(1)  Forces  Intersecting  Within  the  Limits  of  the  Drawing. 
If  the  two  forces  are  not  parallel,  their  lines  of  action  must 
intersect  at  a  point,  which  may  be  taken  as  the  point  of  appli- 
cation of  each  force.    The  two  forces  may  therefore  be  treated 
as  concurrent  forces,  and  their  resultant  may  be  determined 
as  in  §  23. 

(2)  Forces  Intersecting  Outside  of  the  Limits  of  the  Draw- 
ing.    Let  AB  and  BC  (Fig.  7)  represent  in  magnitude  and  direc- 

16 


Art.  1. 


COMPOSITION    OF    NON-CONCURRENT    FORCES. 


17 


tion  two  non-parallei  forces  whose  lines  of  action  ab  and  be 
intersect  outside  of  the  limits  of  the  drawing.  It  is  required 
to  find  their  resultant. 


FIG   7 

Draw  the  force  polygon  ABC  (Fig.  7),  and  connect  the 
points  A  and  C  by  the  closing  line  AC;  then  AC,  acting  as 
shown  by  the  arrow,  represents  in  magnitude  and  direction 
the  resultant  of  the  two  forces  AB  and  BC,  its  line  of  action 
being  as  yet  unknown.  To  determine  this  line  of  action,  an 
auxiliary  construction  is  necessary.  From  any  point  O,  draw 
the  lines  OA,  OB,  and  OC  connecting  the  point  O  with  the 
points  A,  B,  and  C  of  the  force  polygon.  These  lines  represent 
in  magnitude  and  direction  components  into  which  the  forces 
AB  and  BC  may  be  resolved.  Thus,  AB  is  equivalent  to  the 
two  forces  represented  in  magnitude  by  AO  and  OB,  acting 
in  the  directions  shown  by  the  arrows.  Likewise,  BC  is 
equivalent  to  the  two  forces  represented  in  magnitude  and 
direction  by  BO  and  OC.  To  determine  the  lines  of  action  of 
these  components,  start  at  any  point  on  the  line  of  action  ab, 
and  draw  ao  and  ob  parallel  respectively  to  AO  and  OB. 
Prolong  ob  until  it  intersects  the  line  of  action  be;  and  from 
the  intersection  of  ob  and  be,  draw  the  line  oc  parallel  to  OC 
to  intersect  the  line  ao.  The  lines  ao,  ob,  bo,  and  oc  are  then 
the  lines  of  action  of  the  components  AO,  OB,  BO,  and  OC, 
into  which  the  original  forces  have  been  resolved.  Now  the 
forces  represented  by  OB  and  BO  are  equal,  have  the  same 
line  of  action,  and  act  in  opposite  directions;  hence  they 
neutralize  each  other  and  may  be  omitted  from  the  system, 


18  NON-CONCURRENT    FORCES.  Chap.  111. 

thus  leaving  the  two  forces  represented  in  magnitude  and 
direction  by  AO  and  OC  and  in  line  of  action  by  ao  and  oc, 
respectively.  The  resultant  of  these  two  forces,  which  have 
been  shown  equivalent  to  the  given  forces  AB  and  BC,  is  the 
force  represented  in  magnitude  and  direction  by  AC  and  in 
line  of  action  by  ac,  drawn  through  the  intersection  of  ao  and 
oc,  parallel  to  AC. 

The  method  explained  above  is  of  great  importance  and 
should  be  thoroughly  understood  by  the  student ;  as  the  prin- 
ciples employed  in  it  will  be  of  great  use  in  solving  succeeding 
problems. 

32.  Resultant  of  Any  Number  of  Non-concurrent,  Non- 
parallel  Forces.  There  are  two  methods  in  general  use  for 
finding  the  resultant  of  any  number  of  non-concurrent  forces. 
These  methods  embody  the  principles  explained  in  §  31,  and 
are  merely  applications  of  the  method  explained  in  that  sec- 
tion. They  depend  upon  whether  the  given  forces  have  inter- 
sections (i)  inside  of  the  limits  of  the  drawing,  or  (2)  outside 
of  the  limits  of  the  drawing.  The  first  method  to  be  described 
is  limited  in  its  application,  but  permits  of  a  simpler  construc- 
tion for  some  problems.  The  second  method,  however,  is  of 
more  general  use ;  as  it  may  be  employed  for  finding  the 
resultant  of  parallel,  as  well  as  non-parallel,  forces. 

(i)  Forces  Having  Intersections  Within  the  Limits  of  the 
Drawing.  Let  AB,  BC,  CD,  and  DE  (Fig.  8)  represent  in  mag- 
nitude and  direction  a  system  of  four  non-parallel,  non-con- 
current forces ;  and  let  ab,  be,  cd,  and  de,  respectively,  repre- 
sent their  lines  of  action.  It  is  required  to  find  the  resultant 
of  the  system  of  forces. 

Applying  the  method  explained  in  (i)  §  31,  the  resultant  of 
the  forces  represented  by  the  lines  AB  and  BC  (Fig.  8)  is  repre- 
sented in  magnitude  and  direction  by  AC,  acting  as  shown  by 
the  arrow,  and  in  line  of  action  by  ac,  drawn  through  the 
intersection  of  ab  and  be,  parallel  to  AC.  This  resultant  may 
then  be  combined  with  the  force  represented  by  CD,  giving 
as  their  resultant  the  force  represented  in  magnitude  and  direc- 
tion by  AD  and  in  line  of  action  by  ad,  drawn  through  the 


Art.  1. 


COMPOSITION   OF   NON-CONCURRENT   FORCES. 


19 


intersection  of  ac  and  cd,  parallel  to  AD.  In  like  manner,  AD 
may  be  combined  with  the  force  DE,  giving  as  their  resultant 
the  force  represented  in  magnitude  and  direction  by  AE  and 
in  line  of  action  by  ae,  drawn  through  the  intersection  of  ad 
and  de,  parallel  to  AE.  This  last  force  AE  represents  in  mag- 


FIG.   8. 

nitude  and  direction  the  resultant  of  the  given  system  of  forces 
AB,  BC,  CD,  and  DE,  its  line  of  action  being  ae. 

It  should  be  borne  in  mind  that  AE  does  not  represent  the 
magnitude  and  direction  of  any  actual  force.  By  the  resultant 
AE  is  meant  a  force  which,  if  applied,  would  produce  the  same 
effect  upon  the  body  as  the  given  forces. 

(2)  Forces  Having  Intersections  Outside  of  the  Limits  of 
the  Drawing.  Let  AB,  BC,  CD,  DE,  and  EF  (Fig.  9)  represent 


Fia.  9. 


in  magnitude  and  direction  a  system  of  non-concurrent  forces, 
and  let  ab,  be,  cd,  de,  and  ef,  respectively,  represent  their  lines 


20  NON-CONCURRENT    FORCES.  Chap.  III. 

of  action.     It  is  required  to  find  the  resultant  of  the  given 
system  of  forces. 

The  resultant  of  the  given  system  of  forces  may  be  found 
by  applying  the  method  explained  in  (2)  §  31.  Construct  the 
force  polygon  ABCDEF,  and  draw  the  closing  line  AF.  Then 
AF,  acting  in  the  direction  shown  by  the  arrow,  represents  in 
magnitude  and  direction  the  resultant  of  the  given  system  of 
forces.  To  find  its  line  of  action,  assume  any  point  O,  and 
draw  the  lines  OA,  OB,  OC,  OD,  OE,  and  OF.  Then  con- 
struct the  polygon  whose  sides  are  oa,  ob,  oc,  od,  oe,  and  of 
(see  (2)  §  31).  Note  that  the  two  lines,  which  represent  the 
components  into  which  each  force  is  resolved  by  the  lines 
drawn  from  the  point  O  to  the  extremities  of  that  force,  are 
respectively  parallel  to  the  two  lines  which  meet  on  the  line 
of  action  on  that  force.  To  find  the  line  of  action  of  the 
resultant,  prolong  the  extreme  lines  oa  and  of  until  they  inter- 
sect, and  through  the  point  of  intersection  draw  af  parallel  to 
AF.  This  is  the  line  of  action  of  the  required  resultant;  for, 
all  the  components,  into  which  the  given  forces  are  resolved 
by  the  lines  drawn  from  the  point  O  to  the  extremities  of  the 
forces,  are  neutralized  (as  shown  by  the  arrows)  except  the 
two  forces  represented  in  magnitude  and  direction  bv_  AO  and 
OF  and  in  line  of  action  by  ao  and  of,  respectively.  The 
resultant  of  these  two  forces,  which  are  equivalent  to  the  given 
system,  is  found  in  magnitude  by  completing  the  force  triangle 
OAF  and  in  direction  by  making  the  resultant  act  around  the 
force  triangle  in  a  direction  opposite  to  the  other  two  forces. 
Since  the  line  of  action  of  the  resultant  must  be  on  the  line 
of  action  of  each  force,  it  must  act  through  the  common  point 
of  each  line,  viz :  their  point  of  intersection. 
\  33.  Funicular  Polygon.  The  polygon  whose  sides  are  oa, 
Job,  oc,  od,  oe,  and  of  (Fig.  9)  is  called  the  funicular  polygon 
/(also  called  the  equilibrium  polygon). 

Pole.     The  point  O   (Fig.  9)  is  called  the  pole  of  the  force 
polygon. 

Pole  Distance.     The  perpendicular  distance  from  the  pole  to 


Art.  1.  COMPOSITION   OF   NON-CONCURRENT   FORCES.  21 

the  line  representing  the  force  in  the  force  polygon  is  called  the 
pole  distance. 

Rays.  The  lines  OA,  OB,  OC,  OD,  OE,  and  OF  (Fig  9), 
drawn  from  the  pole  O  to  the  extremities  of  the  lines  repre- 
senting the  forces  in  the  force  polygon,  are  called  rays.  The 
rays  terminating  at  the  extremities  of  any  side  of  the  force 
polygon,  represent  in  magnitude  the  two  components  which 
may  replace  the  force  represented  by  that  side. 

Strings.  The  sides  oa,  ob,  oc,  od,  oe,  and  of  of  the  funicular 
polygon  are  called  strings.  The  strings  are  parallel  to  the 
corresponding  rays  of  the  force  polygon,  and  are  the  lines  of 
action  of  the  forces  represented  by  the  rays. 

Referring  to  the  diagram  (Fig.  9),  it  is  seen  that  for  each 
ray  in  the  force  polygon  there  is  a  string  parallel  to  it  in  the 
funicular  polygon;  and  further,  that  the  two  rays  drawn  to 
the  extremities  of  any  force  in  the  force  polygon  are  respect- 
ively parallel  to  the  two  strings  which  intersect  on  the  line  of 
action  of  that  force.  Keeping  in  mind  these  facts  will  greatly 
facilitate  the  construction  of  the  funicular  polygon. 

34.  In  both  cases  given  in  §  32,  the  resultant  has  been 
found  to  be  a  single  force.    This  may  not  always  be  true,  and 
the  simplest  system  that  will  replace  the  given  system  may 
be  a  couple ;  as  will  be  shown  in  the  following  section. 

35.  Resultant  of  Any  Number  of  Non-parallel,  Non-con- 
current Forces. — Resultant  a  Couple.     Upon  determining  the 
resultant  of  a  given  system  of  forces,  it  may  be  found  that  the 
first  and  the  last  sides  of  the  funicular  polygon  are  parallel.   If 
this  is  the  case,  the  constructions  shown  in  §  32  do  not  deter- 
mine the  line  of  action  of  the  resultant.     Referring  to  Fig.  9, 
suppose  the  pole  is  taken  on  the  line  AF;  then  the  first  and 
the    last    strings    of   the    funicular    polygon    are    respectively 
parallel  to  AO  and  OF,  and  are  therefore  parallel  to  each  other. 
In  this  case  the  difficulty  is  avoided  by  taking  the  pole  at  some 
point  not  on  the  closing  line  AF.    There  is,  however,  one  par- 
ticular case  in  which  AO  and  OF  will  be  parallel  no  matter 
where  the  pole  is  taken.     Again  referring  to  Fig.  9,  it  is  seen 
that  AO  and  OF  will  be  parallel  if  the  points  A  and  F  coincide. 


22 


NON-CONCURRENT    FORCES. 


Chap.  III. 


These  two  rays  will  then  represent  equal  and  opposite  forces, 
which  cannot  be  combined  into  a  simpler  system  unless  their 
lines  of  action  coincide.  If  their  lines  of  action  ao  and  of  are 
coincident,  the  two  forces,  being  equal  and  opposite  in  direc- 
tion, neutralize  each  other,  and  their  resultant  is  equal  to  zero. 
If  their  lines  of  action  are  not  coincident,  the  system  reduces 
to  a  couple.  Even  if  the  lines  of  action  of  the  forces  repre- 
sented by  AO  and  OF  are  coincident,  the  forces  may  still  be 
considered  as  a  couple  with  an  arm  equal  to  zero ;  hence  the 
following  proposition :  If  the  force  polygon  for  any  system 
of  forces  closes,  the  resultant  is  a  couple. 

By  changing  the  starting  point  of  the  funicular  polygon, 
the  lines  of  action  of  the  forces  represented  by  AO  and  OF 
will  be  changed ;  and  by  taking  a  new  pole,  either  their  magni- 
tudes, or  directions,  or  both  their  magnitudes  and  directions 
may  be  changed.  Hence,  it  is  seen  that  any  number  of  couples 
may  be  found  which  are  equivalent  to  each  other;  since  they 
are  equivalent  to  the  same  system  of  forces. 

36.  Resultant  of  Any  Number  of  Parallel  Forces.  Let 
AB,  BC,  CD,  DE,  and  EF  (Fig.  10)  represent  in  magnitude 
and  direction  a  system  of  parallel  forces ;  and  let  ab,  be,  cd,  de, 
and  ef  represent  their  lines  of  action,  respectively.  It  is 
required  to  find  the  resultant  of  the  given  system. 

-T---T/V 


IB 


1 

L    d 

e 

b 

b 

°J%* 

"o* 

.-fr' 

J 

v 

cU, 

'    e 

"^s 

/ 

,*^ 

a  > 

' 

a 

f 

• 

R 

FIG.    10. 


Construct  the  force  polygon  ABCDEF,  which  in  this  case 
is  a  straight  line ;  since  the  forces  are  parallel.  Then  AF, 
acting  in  the  direction  shown  by  the  arrow,  represents  in 


Art.  2.  RESOLUTION    OF    NON-CONCURRENT    FORCES.  23 

magnitude  and  direction  the  resultant  of  the  given  system  of 
forces.  To  determine  its  line  of  action,  assume  any  pole  O, 
and  draw  the  rays  OA,  OB,  OC,  OD,  OE,  and  OF.  Then 
construct  the  funicular  polygon  whose  sides  are  oa,  ob,  oc,  od, 
oe,  and  of,  and  prolong  the  extreme  strings  oa  and  of  until 
they  intersect.  Through  this  point  of  intersection,  draw  af 
parallel  to  AF,  which  gives  the  line  of  action  of  the  resultant. 
For,  the  given  system  of  forces  may  be  considered  to  be 
replaced  by  the  forces  represented  in  magnitude  and  direction 
by  AO  and  OF  and  in  line  of  action  by  ao  and  of;  since  the 
other  forces  represented  by  the  rays  neutralize  each  other. 
The  resultant  of  these  two  forces  is  given  in  magnitude  and 
direction  by  the  closing  line  AF  of  the  force  triangle  OAF, 
and  in  line  of  action  by  af,  acting  through  the  intersection  of 
ao  and  of. 

37.  Closing  of  the  Funicular  Polygon.  The  given  system 
of  forces  represented  in  Fig.  9  has  been  shown  to  be  equivalent 
to  the  two  forces  represented  in  magnitude  and  direction  by 
AO  and  OF  and  in  line  of  action  by  ao  and  of,  the  first  and 
last  strings  of  the  funicular  polygon.  In  general  these  lines  of 
action  are  not  parallel,  but  it  may  happen  that  they  are  parallel 
or  that  they  coincide.  In  case  they  coincide,  the  funicular 
polygon  is  said  to  be  closed. 


ART.  2.     RESOLUTION  OF  NON-CONCURRENT  FORCES. 

38.  The  problem  of  resolving  a  given  force  into  two  or 
more    non-concurrent    components    is    indeterminate    unless 
additional  data  are  given  concerning  the  magnitudes  and  the 
lines  of  action  of  the  required  components.     A   given  force 
may  be  resolved  into  three  non-parallel,  non-concurrent  com- 
ponents, or  into  two  parallel  components,  provided  the  lines 
of  action  of  these  two  components  are  given.     For  a  greater 
number  of  components  the  problem  is  indeterminate. 

39.  Resolution  of  a  Force  Into  Three  Non-parallel,  Non- 
concurrent  Components  Having  Known  Lines  of  Action.     Let 


24 


NON-CONCURRENT    FORCES. 


Chap.  III. 


AB  (Fig.  n)  represent  in  magnitude  and  direction  a  given 
force,  and  let  ab  be  its  line  of  action.  It  is  required  to  resolve 
this  force  into  three  components  acting  along  the  lines  cb,  dc, 
and  ad. 


FIG.  11. 

Since  the  given  force  AB  may  be  assumed  to  act  at  any 
point  in  its  line  of  action,  let  its  point  of  application  be  taken  at 
the  intersection  of  ab  and  cb.  Prolong  the  lines  of  action  ad 
and  dc  until  they  intersect,  and  connect  this  point  with  the 
point  of  intersection  of  ab  and  cb.  Resolve  the  given  force 
AB  into  two  components  acting  along  ac  and  cb  (§  27)  ;  then 
AC  and  CB,  acting  as  shown  by  the  arrows,  will  represent  in 
magnitude  and  direction  two  components  of  AB.  In  like  man- 
ner, resolve  the  force  represented  by  AC,  whose  line  of  action 
is  ac,  into  two  components  acting  along  the  lines  ad  and  dc. 
These  two  components  are  given  in  magnitude  and  direction 
by  the  lines  AD  and  DC,  drawn  parallel  respectively  to  ad 
and  dc.  Hence,  since  the  given  force  AB  is  equivalent  to  the 
two  forces  AC  and  CB,  and  AC  is  equivalent  to  the  two  forces 
AD  and  DC;  therefore,  the  force  AB  is  equivalent  to  the 
three  forces  represented  in  magnitude  and  direction  by  AD, 
DC,  and  CB. 

If  the  line  of  action  of  the  given  force  does  not  intersect 
any  of  the  given  lines  of  action  within  the  limits  of  the  draw- 
ing, the  given  force  may  be  replaced  by  two  components,  each 
component  then  resolved  by  the  above  method,  and  the  result- 
ing forces  combined. 

40.  Resolution  of  a  Force  Into  Two  Parallel  Components 
Having  Known  Lines  of  Action.  There  are  two  special  cases 


RESOLUTION    OF    NON-CONCURRENT    FORCES. 


25 


of  this  problem  depending  upon  whether  the  line  of  action  of 
the  given  force  is  (i)  between  the  lines  of  action  of  the  two 
components,  or  (2)  is  outside  of  the  lines  of  action  of  the  two 
components. 

(i)  Line  of  Action  of  Known  Force  Between  the  Lines  of 
Action  of  the  Two  Components.  Let  AB  (Fig.  12)  represent  in 
magnitude  and  direction  the  given  force,  and  let  ab  be  its  line 
of  action ;  also  let  ac  and  cb  be  the  lines  of  action  of  the  two 
parallel  components.  It  is  required  to  find  the  magnitudes  and 
directions  of  the  two  components. 

The  method  explained  in  (2)  §  31  may  be  used  to  find  the 
resultant  of  two  parallel  forces ;  and  conversely,  it  may  be 
used  to  resolve  a  force  into  its  two  components. 


^*  o 


tc' 


(a) 


(b) 


FIG.  12. 


Assume  any  pole  O  (Fig.  12),  and  draw  the  rays  OA  and 
OB.  At  any  point  on  ab,  the  line  of  action  of  the  force  AB, 
draw  the  string  ao  parallel  to  AO,  and  ob  parallel  to  OB.  Pro- 
long the  two  strings  until  they  intersect  ac  and  cb ;  join  these 
points  of  intersection  by  the  string  oc;  and  from  the  pole  O, 
draw  the  ray  OC  parallel  to  the  string  oc,  cutting  the  line  AB 
at  the  point  C.  Then  AC  and  CB,  acting  in  the  directions 
shown  by  the  arrows,  represent  in  magnitude  and  direction 
the  two  components  whose  lines  of  action  are  ac  and  cb, 
respectively.  For,  the  diagram  shown  in  Fig.  12,  b  is  the 
force  polygon  for  the  two  components  AC  and  CB  and  their 
resultant  AB ;  conversely,  AB  is  resolved  into  its  two  com- 
ponents AC  and  CB. 

A  practical  application  of  this  case  is  the  determination  of 


26  NON-CONCURRENT    FORCES.  Chap.   III. 

the  magnitudes  of  the  two  reactions  of  a  simple  beam  loaded 
at  any  point  with  a  single  concentrated  load. 

The  line  oc  is  called  the  closing  string  of  the  funicular  poly- 
gon, and  the  ray  OC  is  called  the  dividing  ray  of  the  force  poly- 
gon ;  since  it  divides  the  given  force  into  its  two  components. 

(2)  Line  of  Action  of  the  Known  Force  Outside  of  the  Lines 
of  Action  of  the  Two  Components.  The  solution  of  this  case  will 
be  left  to  the  student. 


ART.  3.     EQUILIBRIUM  OF  NON-CONCURRENT  FORCES. 

41.  Conditions  for  Equilibrium  of  Non-Concurrent  Forces. 
It  has  been  shown  that  a  system  of  concurrent  forces  is  in 
equilibrium  if  the  force  polygon  closes ;  as  the  resultant  is  then 
equal  to  zero.  In  order  that  a  system  of  non-concurrent  forces 
be  in  equilibrium,  it  is  necessary  that  the  force  polygon  close, 
but  this  single  condition  is  not  sufficient  to  insure  equilibrium. 
For,  referring  to  Fig.  9,  it  is  seen  that  the  given  system  of 
forces  may  be  reduced  to  two  forces  represented  in  magnitude 
and  direction  by  AO  and  OF  and  in  lines  of  action  by  ao  and 
of,  respectively.  For  equilibrium,  these  two  forces  must  be 
equal,  must  have  the  same  line  of  action,  and  must  act  in 
opposite  directions.  It  is  readily  seen  that  the  two  forces  are 
equal  and  act'  in  opposite  directions  if  the  points  A  and  F 
coincide ;  or  in  other  words,  if  the  force  polygon  closes.  In 
order  that  they  have  the  same  line  of  action,  the  first  and  last 
strings,  ao  and  of,  of  the  funicular  polygon  must  coincide. 
From  the  above,  it  is  seen  that  for  equilibrium  of  a  system  of 
non-concurrent  forces : 

(1)  The  force  polygon  must  close. 

(2)  The  funicular  polygon  must  close. 

Therefore,  a  system  of  non-concurrent  forces  is  in  equilib- 
rium if  both  the  force  and  funicular  polygons  close ;    and  con- 
versely, if  both  the  force  and  funicular  polygons  close,  the  sys- 
tem is  in  equilibrium.     For,  if  the  force  polygon  closes,  the 
two  forces  AO  and  OF   (Fig.  9)   are  equal  and  opposite  in 


Art.  3. 


EQUILIBRIUM    OE    NON-CONCURRENT    FORCES. 


27 


direction,  and  if  the  funicular  polygon  closes,  they  have  the 
same  line  of  action  and  neutralize  each  other. 

42.  Use  of  Force  and  Funicular  Polygons  in  Solving  Prob- 
lems in  Equilibrium.     It  has  been  shown  in  Art.  2  that  the 
force    and   the    funicular   polygon    construction    is   especially 
adapted  to  the  solution  of  problems  involving  non-concurrent 
forces.     The  two  conditions  necessary  for  equilibrium  of  non- 
concurrent  forces,  viz. :   that  the  force  polygon  must  close  and 
that  the  funicular  polygon  must  close,  furnish  a  convenient 
graphical  method  for  the  solution  of  problems  in  equilibrium. 
In  order  that  such  problems  may  be  solved,  it  is  necessary  that 
a  sufficient  number  of  forces  be  completely  or  partly  known  to 
permit  of  the  construction  of  the  complete  force  and  funicular 
polygons;  otherwise  the  problem  is  indeterminate. 

The  general  method  of  procedure  is  as  follows:  First  con- 
struct as  much  of  the  force  and  funicular  polygons  as  is  pos- 
sible from  the  given  data ;  then  complete  these  polygons,  keep- 
ing in  mind  the  facts  that  both  polygons  must  close,  and  that 
the  forces  must  act  continuously  around  the  force  polygon. 

The  application  of  the  above  principles  to  the  solution  of 
some  of  the  most  important  cases  arising  in  practice  will  now 
be  given. 

43.  Problem  i.     Parallel  Forces.     Given  a  system  of  par- 
allel forces  in  equilibrium,  all  being  completely  known  except 
two,   these   two  being  known   in   line   of   action   only.      It   is 
required  to  find  the  unknown  elements  of  the  two  forces. 


ATT 
I 
I 
A 


a 

b        bjc 

a/oX 

'"     1°  Ts^ 

,      M 

X    t 

/ 

_     0                          ^-^ 

e 

a 

d 

FIG.  13. 


28  NON-CONCURRENT    FORCES.  Chap.   III. 

Let  the  known  forces  be  represented  by  the  three  loads  AB, 
BC,  and  CD  (Fig.  13),  applied  to  the  beam  along  the  lines  ab, 
be,  and  cd,  and  let  the  unknown  forces  be  the  supporting  forces 
of  the  beam,  EA  and  DE. 

Draw  the  portion  of  the  force  polygon  ABCD  containing 
the  three  known  forces  laid  off  consecutively.  From  the  con- 
dition that  the  force  polygon  must  close,  the  combined  magni- 
tudes of  the  two  supporting  forces  must  be  equal  to  DA  and 
must  act  in  the  direction  from  D  towards  A.  To  determine  the 
magnitude  of  each  supporting  force,  take  any  pole  O,  and 
draw  the  rays  OA,  OB,  OC,  and  OD.  Then,  commencing  at 
any  point  on  ea,  the  line  of  action  of  the  left  supporting  force, 
construct  the  portion  of  the  funicular  polygon  whose  sides  are 
oa,  ob,  oc,  and  od,  drawn  parallel  respectively  to  the  rays  OA, 
OB,  OC,  and  OD.  Prolong  od  until  it  intersects  de,  and  close 
the  funicular  polygon  by  drawing  the  line  oe.  Also,  from  the 
pole  O,  draw  the  ray  OE  parallel  to  the  closing  line  oe,  cutting 
DA  at  E.  Then  EA  and  DE,  acting  in  an  upward  direc- 
tion, represent  in  magnitude  and  direction  the  two  supporting 
forces. 

To  thoroughly  understand  this  solution,  the  student  should 
follow  through  the  constructions  from  the  principle  of  the  tri- 
angle of  forces.  Thus  the  force  AB  is  resolved  into  two  com- 
ponents represented  in  magnitude  and  direction  by  AO  and 
OB  and  in  line  of  action  by  ao  and  ob,  respectively.  In  like 
manner,  the  force  BC  is  resolved  into  the  two  components 
BO  and  OC,  acting  along  the  lines  bo  and  oc;  and  the  force 
CD  is  resolved  into  the  two  components  CO  and  OD,  acting 
along  the  lines  co  and  od,  respectively.  The  two  forces  BO 
and  OB  neutralize  each  other;  since  they  are  equal,  act  in 
opposite  directions,  and  have  the  same  line  of  action.  For  the 
same  reasons,  EO  neutralizes  OE,  and  CO  neutralizes  OC. 
The  three  given  forces  are  therefore  equivalent  to  the  two 
forces  AO  and  OD,  acting  along  the  lines  ao  and  od,  respect- 
ively. Now  in  order  to  have  equilibrium,  the  resultant  of  AO 
and  the  left  supporting  force  EA  must  neutralize  the  resultant 
of  OD  and  the  right  supporting  force  DE.  The  lines  of  action 


Art.  3. 


EQUILIBRIUM   OF    NON-CONCURRENT    FORCES. 


29 


of  these  two  resultants  must  coincide  in  eo  in  order  that  they 
may  neutralize  each  other.  Since  the  resultant  EO  and  the 
two  forces  AO  and  EA  all  intersect  at  a  point,  they  must 
form  a  force  triangle  AOE  (§  23),  of  which  AO  is  the  known 
side.  Then  EA,  acting  in  a  direction  from  E  towards  A,  rep- 
resents the  left  supporting  force  in  magnitude  and  direction. 
In  like  manner,  it  may  be  shown  that  DE  represents  the  right 
supporting  force  in  magnitude  and  direction. 

44.  Problem  2.  Non-parallel  Forces.  Given  a  system  of 
non-parallel  forces  in  equilibrium,  all  completely  known  except 
two.  Of  these  two,  the  line  of  action  of  one  and  the  point  of 
application  of  the  other  are  known.  It  is  required  to  find  the 
unknown  elements  of  the  two  forces. 

Let  the  known  forces  be  the  three  wind  loads  AB,  BC,  and 
CD  (Fig.  14)  acting  on  the  roof  truss,  and  let  the  unknown 
forces  be  the  two  supporting  forces  of  the  truss.  The  line  of 
action  de  of  the  right  supporting  force  is  'given  as  vertical,  and 
the  point  of  application  of  the  left  supporting  force  ea  is  given 
at  the  left  end  of  the  truss. 


FIG.  14. 

Construct  the  portion  of  the  force  polygon  ABCD,  contain- 
ing the  three  known  wind  forces  AB,  BC,  and  CD  laid  off 
consecutively ;  and  from  any  pole  O,  draw  the  rays  OA,  OB, 
OC,  and  OD.  The  side  DE  of  the  force  polygon  must  be  par- 
allel to  the  known  direction  de ;  but  as  its  magnitude  is 
unknown,  as  is  also  the  magnitude  of  the  supporting  force 
EA,  the  polygon  cannot  be  closed.  Since  the  only  known 


30 


NON-CONCURRENT    FORCES. 


'Chap.  III. 


point  in  the  line  of  action  of  the  left  supporting  force  is  its 
point  of  application  at  the  left  end  of  the  truss,  let  the  funicu- 
lar polygon  be  started  at  this  point.  Draw  the  strings  oa,  ob, 
oc,  and  od  parallel  respectively  to  the  rays  OA,  OB,  OC,  and 
OD,  keeping  in  mind  the  fact  that  the  lines  of  action  of  the 
two  components,  into  which  each  force  is  resolved  by  the  rays, 
must  intersect  on  the  line  of  action  of  that  force.  Produce  the 
string  od  until  it  intersects  de,  the  line  of  action  of  the  right 
supporting  force,  and  close  the  funicular  polygon  by  drawing 
the  line  oe.  From  the  pole  O,  draw  the  ray  OE  parallel  to 
oe,  and  from  D  draw  the  line  DE  parallel  to  the  known  direc- 
tion de ;  their  intersection  then  determines  the  point  E  of  the 
force  polygon.  Close  the  force  polygon  by  drawing  the  line 
EA ;  then  DE  and  EA,  acting  in  the  directions  shown  by  the 
arrows,  represent  in  magnitude  and  direction  the  two  support- 
ing forces.  The  line  of  action  of  the  left  supporting  force  is 
found  by  drawing  the  line  ea  parallel  to  EA,  through  its 
known  point  of  application  at  the  left  end  of  the  truss. 

45.  Problem  3.  Non-parallel  Forces.  Given  a  system  of 
non-parallel  forces  in  equilibrium,  all  completely  known  except 
three,  these  three  being  known  in  line  of  action  only.  It  is 
required  to  determine  the  unknown  elements  of  the  three 
forces. 

Let  AB,  BC,  and  CD  (Fig.  15)  represent  the  forces  com- 
pletely known,  and  let  DE,  EF,  and  FA  represent  the  three 
forces  which  are  known  in  line  of  action  only. 


FIG.  15. 


Since  the  resultant  of  any  two  forces  must  pass  through 


Art.  3.  EQUILIBRIUM    OF    NON-CONCURRENT    FORCES.  31 

their  point  of  intersection,  let  two  of  the  unknown  forces, 
whose  lines  of  action  are  represented  by  ef  and  fa,  be  replaced 
by  their  resultant  acting  through  their  point  of  intersection. 
The  .problem  then  becomes  identical  with  Problem  2,  de  being 
the  line  of  action  of  one  unknown  force;  and  the  intersection 
of  ef  and  fa,  the  point  of  application  of  the  other  unknown 
force.  The  construction  of  Problem  2  having  been  made,  and 
the  resultant  EA  of  the  two  forces  found,  the  magnitudes  of 
these  forces,  whose  lines  of  action  are  represented  by  ef  and  fa, 
may  be  found  by  resolving  this  resultant  into  two  components 
parallel  to  these  known  lines  of  action.  In  Fig.  15,  EA  repre- 
sents this  resultant  in  magnitude  and  direction,  and  EF  and 
FA  represent  the  two  components  into  which  it  is  resolved. 
Since  the  force  DE  is  given  by  the  construction  of  Problem  2, 
the  three  unknown  forces  are  therefore  represented  in  magni- 
tude and  direction  by  DE,  EF,  and  FA. 

In  these  problems  and  in  those  given  in  the  preceding  sec- 
tions, it  should  particularly  be  noted  that  the  forces  should  be 
used  in  such  an  order  that  those  which  are  completely  known 
are  consecutive. 

46.  Inaccessible  Points  of  Intersection.  In  Problem  3  and 
in  other  problems,  it  may  happen  that  the  point  of  intersection 
of  the  two  forces  falls  outside  of  the  limits  of  the  drawing. 
When  this  is  the  case,  and  when  it  is  required  to  draw  a  line 
from  a  given  point  through  the  point  of  intersection  of  two 
forces  whose  lines  of  action  intersect  outside  of  the  limits  of 
the  drawing,  the  following  geometrical  construction  may  be 
employed: 

Let  XX  and  YY  (Fig.  16)  be  two 
lines  intersecting  outside  of  the 
limits  of  the  drawing.  It  is  re- 
quired to  draw  a  line  through  their 
point  of  intersection  from  a  given 
point  P. 

Draw  any  line  PA  through  the 
point  P  to  intersect  the  lines  XX 
and  YY  at  the  points  A  and  B,  FlG  16 


32  NON-CONCURRENT    FORCES.  Chap.  III. 

respectively.  Also,  draw  the  lines  AC  and  PC  intersecting 
on  YY  to  form  the  triangle  ACP.  From  any  point  A'  on 
XX,  draw  the  lines  A'P'  and  A'C',  parallel  respectively  to 
AP  and  AC,  and  from  C'  draw  the  line  C'P'  parallel  to 
CP  to  intersect  A'P'  at  P'.  Then  PP',  prolonged,  will 
pass  through  the  point  of  intersection  of  XX  and  YY.  For, 
since  the  triangles  ABC  and  BCP  are  similar  to  the  triangles 
A'B'C'  and  B'C'P,  respectively,  therefore  AB  :  A'B'  : : 
CB  :  C'B',  and  BP  :  B'P'  : :  CB  :  C'B',  which  gives  AB  :  A'B'  : : 
BP  :  B'P',  thus  proving  that  the  three  lines  XX,  YY,  and  PP' 
meet  in  a  point. 

47.  Problems,  (i).  A  rigid  beam  16  feet  in  length  rests 
horizontally  upon  supports  at  its  ends  and  carries  the  follow- 
ing loads:  its  own  weight  of  200  Ibs.,  acting  at  its  center, 
and  three  loads  of  150  Ibs.,  80  Ibs.,  and  120  Ibs.,  acting  at 
points  which  are  distant  4  ft.,  i'o  ft.,  and  13  ft.,  respectively, 
from  the  left  support.  Find  the  upward  pressures,  or  reac- 
tions, at  the  supports. 

(2).    Given  the  roof  truss  and  wind  loads  as  shown  (Fig. 

17).  The  truss  is  fixed  to  the 
wall  at  the  right  end,  and  is 
supported  on  rollers  at  the  left. 
It  is  required  to  find  the  mag- 
nitude and  line  of  action  of  the 
right  reaction  and  the  magni- 
P~  17.  tude  of  the  left.  (The  left  re- 

action  must   be   vertical;    since 

the  left  end  of  the  truss  is  on  rollers  and  can  have  no  hori- 
zontal component.) 

(3).  A  uniform  bar  24  inches  long,  weighing  20  Ibs.,  has 
one  end  resting  against  a  smooth  wall,  and  is  supported  on 
a  smooth  peg,  which  is  13  inches  from  the  wall.  The  bar  is 
held  in  equilibrium  at  an  angle  of  60  degrees  with  the  ver- 
tical by  a  weight  W  suspended  from  the  free  end  of  the  bar. 
It  is  required  to  find  the  weight  W,  and  the  pressures  against 
the  bar  by  the  wall  and  the  peg. 


Art  3. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES. 


48.  Special  Method.  If  any  system  of  forces  in  equilib- 
rium is  divided  into  two  groups,  the  resultants  of  the  two 
groups  must  be  equal,  must  act  in  the  opposite  directions,  and 
must  have  the  same  line  of  action.  These  facts  suggest  a 
special  method  for  solving  certain  problems,  which  in  some 
cases  is  simpler  than  the  general  method  of  constructing  the 
force  and  funicular  polygons. 

Problem.  Let  AB  (Fig.  18)  represent  a  force  which  is  com- 
pletely known;  and  let  BC,  CD,  and  DA  be  known  in  line 
of  action  only,  the  lines  of  action  of  the  four  forces  being  ab, 
be,  cd,  and  da. 


FIG.  is. 

The  resultant  of  the  forces  whose  lines  of  action  are  ab 
and  be  must  pass  through  their  point  of  intersection ;  also,  the 
resultant  of  the  forces  whose  lines  of  action  are  cd  and  da 
must  pass  through  the  intersection  of  cd  and  da.  For  equi- 
librium, these  two  resultants  must  be  equal,  must  have  oppo- 
site directions,  and  must  have  the  same  line  of  action;  hence 
each  must  act  through  the  line  ac.  Draw  AB  representing  the 
magnitude  and  direction  of  the  known  force,  and  from  A  and 
B,  the  extremities  of  the  line  AB,  draw  lines  parallel  respect- 
ively to  ac  and  be.  Then  BC  represents  the  magnitude  and 
direction  of  the  force  whose  line  of  action  is  be;  and  AC,  act- 
ing in  the  direction  from  A  towards  C,  represents  the  magni- 
tude and  direction  of  the  resultant  of  AB  and  BC.  But  for 
equilibrium,  CA,  acting  in  the  opposite  direction,  i.  e.,  from  C 
towards  A,  must  represent  the  resultant  of  the  forces  acting 
along  the  lines  cd  and  da ;  hence  these  forces  are  represented 
in  magnitude  and  direction  by  CD  and  DA,  drawn  from  the 
points  C  and  A,  parallel  respectively  to  cd  and  da. 


34 


NON-CONCURRENT    FORCES. 


Chap.  III. 


Problem.  A  beam  with  rounded  ends  has  one  end  resting 
against  a  smooth  wall  and  the  other  end  on  a  smooth  floor. 
The  weight  of  the  beam  is  60  Ibs.  acting  through  its  center. 
Neglecting  the  friction  of  the  beam  against  the  wall  and  floor, 
what  force  applied  horizontally  at  the  lower  end  of  the  beam 
would  support  it  at  an  angle  of  60  degrees  with  the  hori- 
zontal? 


ART.  4.     SPECIAL  CONSTRUCTIONS  FOR  FUNICULAR  POLYGONS. 


49.  Relation  Between  Different  Funicular  Polygons  for 
the  Same  Forces.  It  will  now  be  shown  that  if  two  funicular 
polygons  are  drawn,  using  the  same  forces  and  force  polygons 
but  different  poles,  the  intersections  of  corresponding  strings 
of  these  funicular  polygons  will  meet  on  a  straight  line  which 
is  parallel  to  the  line  joining  the  two  poles. 

Let  AB,  BC,  and  CD  (Fig.  19)  represent  in  magnitude  and 
direction  a  system  of  forces,  and  let  ab,  be,  and  cd,  respect- 
ively, be  their  lines  of  action. 


FIG.   19. 

Draw  the  force  polygon  ABCD  for  the  given  system  of 
forces,  and  with  the  pole  O,  draw  the  rays  OA,  OB,  OC,  and 
OD ;  also,  with  any  other  pole  O',  draw  the  rays  O'A,  O'B, 
O'C,  and  O'D.  Construct  the  funicular  polygons  whose  sides 
are  oa,  ob,  oc,  and  od,  having  the  strings  respectively  parallel 
to  the  rays  OA,  OB,  OC,  and  OD ;  also,  construct  the  funicu- 
lar polygon  whose  sides  are  o'a,  o'b,  o'c,  and  o'd,  having  the 


Art. 


FUNICULAR  POLYGON   THROUGH  TWO   POINTS. 


35 


strings  respectively  parallel  to  the  rays  O'A,  O'B,  O'C,  and 
O'D.  Produce  the  strings  oa  and  o'a  until  they  intersect.  In 
like  manner,  produce  the  other  corresponding  strings  until 
each  pair  intersects.  Draw  the  line  XX  through  these  points 
of  intersection;  then  XX  will  be  parallel  to  OO',  the  line 
joining  the  two  poles.  For,  the  quadrilaterals  whose  sides  are 
AB,  OA,  OO',  O'B ;  and  ab,  oa,  XX,  o'b  have  by  construc- 
tion three  sides  and  two  diagonals  respectively  parallel  each 
to  each ;  AB  parallel  to  ab,  OA  parallel  to  oa,  O'B  parallel  to 
o'b,  O'A  parallel  to  o'a,  and  OB  parallel  to  ob:  hence  the 
fourth  sides  OO'  and  XX  are  also  parallel.  The  same  relation 
may  be  proved  for  the  quadrilaterals  whose  sides  are  OB,  BC, 
CO',  OO' ;  and  ob,  be,  co',  XX,  and  so  on. 

This  relation  between  the  two  funicular  polygons  is  em- 
ployed for  drawing  the  line  of  pressure  of  an  arch. 

50.  To  Draw  a  Funicular  Polygon  Through  Two  Given 
Points.  Let  AB,  BC,  and  CD  (Fig.  20)  represent  in  magni- 
tude and  direction  a  system  of  forces  acting  on  the  beam ;  and 
let  ab,  be,  cd,  respectively,  be  their  lines  of  action.  It  is 
required  to  draw  a  funicular  polygon  passing  through  the  two 
points  M  and  N,  which  points  are  on  the  lines  of  action  of 
the  two  reactions. 

A 


a  e 


FIG.  20.     FUNICULAR  POLYGON  THROUGH  Two  POINTS. 

Assume  any  pole  O',  and  draw  the  rays  O'A,  O'B,  O'C, 
and  O'D.  Construct  the  funicular  polygon  whose  sides  are 
o'a,  o'b,  o'c,  and  o'd,  and  close  the  polygon  by  drawing  the 


36 


NON-CONCURRENT    FORCES. 


Chap.  III. 


closing  line  o'e.  From  the  pole  O',  draw  the  dividing  ray 
O'E,  cutting  AD  at  the  point  E.  Then  DE  and  EA  represent 
the  two  reactions  whose  lines  of  action  are  de  and  ea,  respect- 
ively. Now  the  point  E  will  remain  fixed,  no  matter  where 
the  pole  is  taken;  since  the  two  reactions  must  be  the  same 
for  the  given  system  of  forces.  From  E  draw  the  line  EO 
parallel  to  the  line  connecting  the  two  given  points  M  and  N. 
Then  the  new  pole '  O,  for  the  funicular  polygon  passing 
through  the  two  given  points,  must  be  on  this  line;  since  OE 
is  the  dividing  ray  of  the  force  polygon,  drawn  parallel  to  the 
closing  string  MN.  With  the  new  pole  O  (at  any  point  on 
OE),  commence  at  M,  and  draw  the  funicular  polygon  whose 
sides  are  oa,  ob,  oc,  od,  and  oe,  which  must  pass  through  the 
given  points  M  and  N. 

51.  To  Draw  a  Funicular  Polygon  Through  Three  Given 
Points.  Let  AB,  BC,  CD,  and  DE  (Fig.  21)  represent  in 
magnitude  and  direction  a  system  of  forces  acting  on  the 
beam ;  and  let  ab,  be,  cd,  and  de,  respectively,  be  their  lines 
of  action.  It  is  required  to  draw  a  funicular  polygon  passing 
through  the  three  given  points  a,  b,  and  c. 


C' 


FIG.  21.     FUNICULAR  POLYGON  THROUGH  THREE  POINTS. 

Assume  a  system  of  forces  to  be  acting  upon  a  beam  of 
such  a  length  that  the  two  reactions  will  pass  through  the 
two  outside  points.  This  length  is  determined  by  drawing 
lines  through  the  two  outside  points  parallel  to  the  equilib- 
rant  of  the  given  forces. 


Art.  4.    FUNICULAR  POLYGON  THROUGH  THREE  POINTS.       37 

Take  any  pole  O',  and  draw  the  rays  O'A,  O'B,  O'C, 
O'D,  and  O'E.  Commencing  at  a,  construct  the  funicular 
polygon  a  b'  c',  and  close  the  polygon  by  drawing  the  closing 
line  ac'.  From  the  pole  O',  draw  the  dividing  ray  O'C',  cut- 
ting EA,  the  equilibrant  of  the  given  forces,  at  C'.  Then 
EC'  and  C'A  represent  the  two  reactions  whose  lines  of  action 
are  ec'  and  c'a,  and  which  pass  through  c  and  a,  respectively. 
Now  draw  the  line  DA,  which  represents  the  equilibrant  of 
the  three  forces  to  the  left  of  b.  Through  b,  draw  the  line 
bb'  parallel  to  DA.  Connect  the  points  a  and  b',  and  the 
points  a  and  b  by  the  lines  ab'  and  ab,  which  are  the  closing 
strings  of  the  funicular  polygons  for  the  forces  to  the  left  of  b. 
Through  O',  draw  the  dividing  ray  O'B'  parallel  to  ab',  cut- 
ting DA  at  B'.  Then  DB'  and  B'A  represent  the  reactions  of 
the  forces  to  the  left  of  b,  acting  through  the  points  b  and  a, 
respectively.  Point  C'  is  common  to  all  force  polygons  for 
the  given  system  of  forces,  and  point  B'  is  common  to  all  force 
polygons  for  the  forces  to  the  left  of  b.  From  the  points  C' 
and  B',  draw  lines  parallel  respectively  to  ac  (the  line  con- 
necting two  of  the  given  points  a  and  c)  and  ab,  intersecting 
at  O.  Then  these  lines  are  the  dividing  rays  corresponding 
:o  the  closing  strings  of  the  required  funicular  polygon,  and 
:heir  intersection  O  will  determine  the  pole  for  this  polygon. 
With  this  pole  O,  commence  at  a,  and  draw  the  required  funic- 
ular polygon,  which  must  pass  through  the  three  given  points 
a,  b,  and  c.  For,  if  the  pole  O  is  on  the  line  C'O,  the  funicular 
polygon  must  pass  through  the  points  a  and  c;  and  if  it  is 
on  true  line  B'O,  the  funicular  polygon  must  pass  through  the 
points  a  and  b. 

This  construction  and  that  shown  in  §  50  may  be  used  to 
determine  whether  or  not  the  line  of  pressure  in  a  masonry 
arch  remains  within  the  middle  third  at  all  joints. 


CHAPTER  IV. 
MOMENTS.' 

This  chapter  is  divided  into  two  articles,  viz.:  Art.  i, 
Moments  of  Forces  and  of  Couples,  and  Art.  2,  Graphic 
Moments.  The  first  article  treats  of  the  general  principles  of 
the  moments  of  forces  and  of  couples ;  and  the  second  treats 
of  the  graphic  determination  of  moments. 

ART.  i.     MOMENTS  OF  FORCES  AND  OF  COUPLES. 

52.  Moment  of  a  Force.  The  moment  of  a  force  about 
any  point  is  the  product  of  the  magnitude  of  the  force  into 
the  perpendicular  distance  of  its  line  of  action  from  the  given 
point.  The  point  about  which  the  moment  is  taken  is  called 
the  origin  (or  center)  of  moments;  and  the  perpendicular 
distance  from  the  origin  to  the  line  of  action  of  force  is  called 
the  arm  of  the  force. 


a 

0  SO' 

Moment=+Pa  Mornent=-PcT 

(a)  (b)  &  0  C 

FIG.  22.  FIG.  23. 

53.  Positive  and  Negative  Moments.  Rotation  may  be  in 
either  of  two  opposite  directions,  viz. :  in  the  direction  of  the 
hands  of  a  clock,  or  opposite  to  the  direction  of  the  hands  of  a 

38 


A**'  1-  MOMENT    AREA.  39 

clock.  The  first  will  be  called  positive ;  and  the  second,  nega- 
tive. The  sign  of  the  moment  of  the  force  is  taken  the  same 
as  that  of  the  direction  of  the  rotation  it  tends  to  produce. 
Thus  the  moment  of  the  force  P  (Fig.  22,  a)  about  the  point 
O  is  equal  to  +Pa,  and  the  moment  of  P'  (Fig.  22,  b)  about 
the  point  O'  is  equal  to  — P'a'. 

54.  Moment  Area.     The  moment  of  a  force  may  be  repre- 
sented by  double  the  area  of  a  triangle,  the  vertex  of  which  is 
at  the  origin  of  moments  and  the  base  a  length  in  the  line  of 
action  of  the  force  equal  to  the  magnitude  of  the  force.    Thus 
the  moment  of  the  force  AB   (Fig.  23)  about  the  point  O  is 
numerically   represented   by   twice   the   area   of  the   triangle 
OAB,  which  is  equivalent  to  the  area  of  the  rectangle  ABCD. 

55.  Transformation  of  Moment  Area.     In  comparing  the 

moments  of  forces,  it  is  often  conven- 

A'  B'  ient   to    transform   their   moment   areas 

into  equivalent  areas  having  a  common 


B 


base.     The  moments  are  then  to  each 


other  as  the  altitudes  of  their  respect- 


1    „ 

,  ive  moment  areas.  Thus  in  Fig.  24,  let 

FIG  24  the  moment  area  represented  by  ABCD 

be  transformed  into  an  equivalent  area 
having  its  base  equal  to  DC'. 

Connect  A  and  C',  and  from  C  draw  CA'  parallel  to  C'A, 
and  prolong  it  until  it  intersects  DA,  prolonged,  at  the  point 
A'.  Complete  the  rectangle  A'B'C'D,  which  is  the  required 
moment  area  equivalent  to  ABCD.  For,  since  the  triangles 
ADC'  and  A'DC  are  similar,  AD  :A'D :  :DC'  :DC;  hence, 
ADXDC=A'DXDC',  or  in  other  words,  the  area  ABCD= 
the  area  A'B'C'D. 

56.  Moment  of  the  Resultant  of  Two  Concurrent  Forces. 
Proposition.  The  moment  of  the  resultant  of  two  concurrent 
forces  about  any  point  in  their  plane  is  equal  to  the  algebraic 
sum  of  their  separate  moments  about  the  same  point. 

Let  DA  (Fig.  25)  represent  the  resultant  of  the  two  con- 
current forces  BA  and  CA  whose  lines  of  action  intersect 
at  A,  and  let  O  be  the  origin  of  moments.  It  is  required  to 


OF  THE 

UNIVERSITY 


40  MOMENTS.  Chap.  IV. 

prove  that  the  moment  of  DA  about  O  is  equal  to  the  alge- 
braic sum  of  the  separate  moments  of  BA  and  CA  about  the 
same  point. 

Connect  the  points  O  and  A  by  the  line  OA,  and  from  O, 
draw  OE  perpendicular  to  OA.    Draw  the  lines  DE,  CF,  and 
BG  parallel  to  OA,  and  join  the  points 
O   and   B,    O   and   C,   and   O  -and   D 
by   the   lines   OB,    OC,   and    OD,    re- 
spectively.   Then    (§    54),   the   moment 
of  BA  about  O  is  equal  to  twice  the 
area  of  the  triangle  OBA  =  +OAX 
OG.     Likewise,    the    moment    of    CA 
FIG.  25.  about  O  is  equal  to  twice  the  area  of 

the  triangle  OCA  =  +OAXOF;  and  the  moment  of  DA 
about  O  is  equal  to  twice  the  area  of  the  triangle  ODA  = 
+OAXOE.  But  OE  =  OF  +  FE,  or  since  FE  =  OG  by 
construction,  then  OE  =  OF  f  OG.  Therefore,  OA  X  OE  = 
OA  (OF  +  OG),  or  the  moment  of  DA  about  the  point  O  is 
equal  to  the  algebraic  sum  of  the  moments  of  BA  and  CA 
about  the  same  point. 

57.  Moment  of  the  Resultant  of  Any  Number  of  Concur- 
rent Forces.     Proposition.     The  moment  of  the  resultant  of 
any   number  of  concurrent   forces   about   any   point   in   their 
plane  is  equal  to  the  algebraic  sum  of  their  separate  moments 
about  the  same  point. 

The  proof  of  this  proposition  follows  directly  from  that 
given  in  §  56,  and  is  simply  an-  extension  of  that  proof. 

58.  Moment  of  a  Couple.     The  moment  of  a  couple  about 
any  point  in  the  plane  of  the  couple  is  equal  to  the  algebraic 
sum  of  the  moments  of  the  two  forces  composing  the  couple 
about  the  same  point. 


It  will  be  shown  that  the  moment  of  a  couple  is 
equal  to  the  product  of  one  of  the  forces  into  the 
perpendicular  distance  between  the  lines  of  action  of  ^    P 
the  forces. 

Let  O  (Fig.  26)  be  the  origin  of  moments,  and         p- 


Q 


let  P  and  P'  be  the  two  equal  forces  of  the  couple.        FlG  26 


Art.  1. 


MOMEXT  OF  THE  RESULTANT  OF  FORCES. 


41 


Then  the  moment  of  the  couple  is  equal  to  —  P'  (a  +  b)  +  Pb 
=  —  P'a  (since  P  is  equal  to  P'),  or  the  moment  of  the  couple 
is  equal  to  the  product  of  one  of  the  forces  into  the  perpendicular 
distance  between  the  two  forces.  Since  O  is  any  point  in  the 
plane  of  the  couple,  it  is  evident  that  the  moment  of  the  couple  is 
independent  of  the  origin  of  moments. 

59.  Moment  of  the  Resultant  of  Any  System  of  Forces. 
Proposition.  The  moment  of  the  resultant  force,  or  resultant 
couple,  of  any  system  of  coplanar  forces  about  any  point  in 
their  plane  is  equal  to  the  algebraic  sum  of  the  separate 
moments  of  the  forces  composing  the  system  about  the  same 
point. 

Let  AB,  BC,  CD,  and  DE  (Fig.  27)  represent  in  magnitude 
and  direction  any  system  of  forces,  and  let  ab,  be,  cd,.and  de 
represent  their  lines  of  action,  respectively. 


A 


FIG.  27. 

Assume  any  pole  O,  and  draw  the  rays  OA,  OB,  OC,  OD, 
and  OE.  Also,  draw  the  funicular  polygon  whose  sides  are 
oa,  ob,  oc,  od,  and  oe.  Now,  AB  may  be  replaced  by  AO  and 
OB,  acting  along  ao  and  ob ;  BC  may  be  replaced  by  BO  and 
OC,  acting  along  bo  and  oc ;  CD  may  be  replaced  by  CO  and 
OD,  acting  along  co  and  od ;  and  DE  may  be'replaced  by  DO 
and  OE,  acting  along  do  and  oe.  (By  AB,  BC,  etc.,  are  meant 
the  forces  represented  in  magnitude  and  direction  by  AB,  BC, 
etc.).  Now  (§  56),  no  matter  where  the  origin  of  moments  is 
taken : 


42  MOMENTS.  Chap.  IV. 

Moment  of  AB  =  moment  of  AO  +  moment  of  OB ; 
"  BC  =  "  "  BO  +  "  "  OC; 
"  CD  =  "  "  CO  +  "  "  OD; 
"  DE  =^  "  "  DO  +  "  "  OE. 

Since  the  forces  represented  by  OB  and  BO  are  equal  in 
magnitude,  opposite  in  direction,  and  have  the  same  line  of 
action,  their  moments  are  equal  but  have  opposite  signs.  In 
like  manner,  the  moments  of  OC  and  CO,  and  of  OD  and  DO 
are  equal  but  have  opposite  signs.  The  addition  of  the  above 
four  equations  shows  that  the  sum  of  the  moments  of  AB, 
BC,  CD,  and  DE  is  equal  to  the  sum  of  the  moments  of  AO 
and  OE.  Now  the  resultant  of  the  given  system  of  forces 
may  be  either  a  resultant  force  or  a  resultant  couple  (§32  and 
§  35).-  In  the  former  case,  the  resultant  of  the  system  is  the 
resultant  of  AO  and  OE,  which  is  AE;  and  its  moment  is 
equal  to  the  algebraic  sum  of  their  moments,  or  to  the  moment 
of  AE  (§  56).  In  the  latter  case,  which  occurs  only  when  E 
coincides  with  A,  the  resultant  is  composed  of  the  forces  AO 
and  OE.  These  forces  are  equal,  act  in  opposite  directions, 
and  form  a  couple  whose  moment  is  equal  to  the  algebraic 
sum  of  the  moments  of  AO  and.OE.  The  proposition  is  there- 
fore true  in  either  case. 

It  should  be  noticed  that  the  proof  here  given  applies  to 
any  system  of  forces,  whether  parallel  or  non-parallel. 

60.  Moment   of   a   System   of    Forces.      Definition.     The 
moment  of  a  system  of  forces  is  equal  to  the  algebraic  sum  of 
the  moments  of  the  forces  composing  the  system. 

61.  Condition  of  Equilibrium.        Proposition.     If  a  given 
system  of  forces  is  in  equilibrium,  the  algebraic  sum  of  their 
moments  about  every  origin  must  be  equal  to  zero.     For,  in 
order  that  the  given  system  of  forces  shown  in  Fig.  27  be  in 
equilibrium,  AO  and  OE  must  be  equal,  must  act  in  opposite 
directions,  and  must  have  the  same  line  of  action.     The  sum 
of  their  moments,  which  is  equal  to  the  sum  of  the  moments 
of  the  given  system  of  forces,  is  therefore  equal  to  zero.    The 
converse    of   this    proposition,    if   the    algebraic    sum    of    the 
moments  about  every  origin  is  equal  to  zero,  the  system  of 


Art.  2.  GRAPHIC    MOMENTS.  43 

forces  is  in  equilibrium,  is  also  true.  For,  if  the  sum  of  their 
moments  is  not  equal  to  zero,  the  system  must  have  either  a 
resultant  force  or  a  resultant  couple.  If  the  resultant  is  a 
force,  then  its  moment  is  not  equal  to  zero  unless  the  origin 
is  taken  on  its  line  of  action ;  and  if  the  resultant  is  a  couple, 
then  its  moment  is  not  equal  to  zero  for  any  origin,  no  matter 
where  taken.  Therefore,  if  the  sum  of  the  moments  about 
every  origin  is  equal  to  zero,  the  system  has  neither  a  resultant 
force  nor  a  resultant  couple,  and  must  be  in  equilibrium. 


ART.  2.     GRAPHIC  MOMENTS. 

62.     Graphic  Representation  of  the  Moment  of  a  Force. 
Proposition.     If  through  any  point  in  the  space  diagram  a  line 

is    drawn    parallel    to    a    given 
force,  the  distance  intercepted 
upon    this    line    by    the    two 
*x  strings    corresponding    to    the 


a 


h 


-7-^*0    components     into     which     the 

•  ~  given  force  is  resolved  by  the 

rays  multiplied  by  the  pole  dis- 
tance of  the   force   is   equal   to 
the  moment  of  the  force  about  the  given  point. 

Let  AB  (Fig.  28)  represent  in  magnitude  and  direction  a 
force  whose  line  of  action  is  ab.  Also,  let  P  be  the  origin  of 
moments,  and  H  the  pole  distance  of  the  given  force.  From 
the  pole  O,  draw  the  rays  OA  and  OB,  and  from  any  point  on 
ab,  draw  the  strings  oa  and  ob  parallel  respectively  to  OA 
and  OB.  Also,  from  the  origin  P,  draw  a  line  parallel  to  the 
line  of  action  of  the  given  force  AB,  intersecting  these  strings 
and  cutting  off  the  intercept  y.  It  is  required  to  prove  that 
the  moment  of  the  given  force  P  is  equal  to  H  multiplied  by  y. 
Let  h  be  the  perpendicular  distance  from  P  to  ab.  Then 
the  moment  of  AB  about  P  is  equal  to  — AB  X  h.  From  sim- 
ilar triangles,  H  :  h  : :  AB  :  y ;  therefore  H  X  y  =  AB  X  h, 
which  proves  the  proposition. 


44  MOMENTS.  Chap.  IV. 

It  should  be  noticed  that  the  intercept  always  represents 
a  distance  and  is  measured  to  the  same  scale  as  the  distances 
in  the  space  diagram,  and  that  the  pole  distance  always  repre- 
sents a  force  magnitude  and  is  measured  to  the  same  scale 
as  the  forces  in  the  force  diagram. 

63.  Moment  of  Any  System  of  Forces.  Proposition.  The 
moment  of  any  system  of  coplanar  forces  about  any  origin  in 
the  plane  is  equal  to  the  distance  intercepted,  on  a  line  drawn 
through  the  origin  of  moments  parallel  to  fhe  resultant  of  all 
the  forces,  by  the  strings  which  meet  upon  the  line  of  action 
of  the  resultant,  multiplied  by  the  pole  distance  of  the 
resultant. 

Let  ABCDE  (Fig.  29)  be  the  force  polygon,  and  let  the 
polygon  whose  sides  are  oa,  ob,  oc,  od,  and  oe  be  the  funicular 
polygon  for  the  given  system  of  forces  AB,  BC,  CD,  and  DE. 
Also,  let  R  represent  the  resultant  of  the  system,  let  H  be  the 
pole  distance  of  this  resultant,  and  y  the  distance  intercepted, 
on  a  line  drawn  through  the  origin  of  moments  P  parallel  to 
the  resultant,  by  the  strings  meeting  on  the  line  of  action  of 
the  resultant.  It  is  required  to  prove  that  the  moment  M  of 
the  given  system  of  forces  is  equal  to  H  X  y- 


The  moment  of  the  resultant  is  equal  to  the  moment  of  the 
given  system  of. forces  (§  59).  Then  the  moment  M  of  the 
given  system  is  equal  to  R  X  h,  where  h  is  the  perpendicular 
distance  from  the  origin  of  moments  to  the  line  of  action  of 


Art. 


MOMENT  OF  PARALLEL  FORCES. 


45 


the  resultant.  From  similar  triangles,  H  :  h  : :  R  :  y ;  therefore, 
H  X  y  =  R  X  h,  which  proves  the  proposition. 

Referring  to  the  space  diagram  (Fig.  29),  it  is  seen  that 
the  resultant  R  tends  to  produce  rotation  in  the  direction  of 
the  hands  of  a  clock,  and  is  positive.  The  magnitude  of  this 
moment  decreases  as  the  origin  approaches  the  line  of  action 
of  the  resultant,  until  it  becomes  zero  when  on  the  line  of 
action  of  R.  If  the  origin  passes  through  this  line  of  action, 
the  moment  is  negative. 

64.  Moment  of  Parallel  Forces.  The  method  explained 
in  §  62  and  §  63  is  especially  useful  when  it  is  desired  to  find 
the  moment  of  any,  or  all,  of  a  system  of  parallel  forces.  For 
example,  it  is  required  to  find  the  moment  at  any  point  in  a 
beam  caused  by  several  loads  on  the  beam. 

Let  M-N  (Fig.  30)  be  a  simple  beam  loaded  with  the  loads 
AB,  BC,  CD,  and  DE.  It  is  required  to  find  the  moment  at 
any  point  P  due  to  the  forces  to  the  left  of  P. 

A 
alb    blc     eld     die 


R. 


FIG.  30. 

Construct  the  force  and  funicular  polygons  for  the  given 
system  of  forces,  as  shown  in  Fig.  30,  and  determine  the  magni- 
tudes of  the  reactions  R!  and  R2.  Produce  each  string  of  the 
funicular  polygon  until  it  intersects  a  line  drawn  through  P 
parallel  to  the  resultant  of  all  the  forces.  Let  H  be  the  pole 
distance  for  the  given  forces,  which  in  this  case  is  the  same 
for  all  the  forces;  since  the  force  polygon  is  a  straight  line. 
Then  (§  62),  considering  the  forces  to  the  left  of  P: 


46  MOMENTS.  Chap.  IV. 

Moment  of  Rt  about  P  =  +H   X   mr; 
"  AB     "       P  =  — H   X   mn ; 
«-  BC     "       P  =  — H   X   no; 
"  CD     "       P  =  — H  X  op. 

The  addition  of  these  equations  gives  the  sum  of  the  moments 
of  Rlf  AB,  BC,  and  CD  =  H  (+  mr  —  mn  —  no  —  op)  =  +  H 
X  y,  which  is  equal  to  the  moment  in  the  beam  at  the  point  P. 
In  like  manner,  it  may  be  shown  that  the  moment  of  the  forces 
to  the  right  of  P  is  equal  to  —  H  X  y.  /  Since  P  is  any  point 
along  the  beam,  it  is  seen  that  the  moment  at  any  point  is 
equal  to  the  ordinate  of  the  funicular  polygon,  cut  off  by  a  line 
through  the  given  point  parallel  to  the  resultant  of  all  the 
jforces,  multiplied  by  the  pole  distance.  \ 

The  summation  of  the  moments  of  all  the  forces  to  the  left 
of  any  point  in  the  beam  about  that  point  is  called  the  bending 
moment. 

The  ordinate  represents  a  distance  and  is  measured  to  the 
same  scale  as  the  beam,  while  the  pole  distance  represents  a 
force  and  is  measured  to  the  same  scale  as  the  forces  in  the 
force  polygon. 

65.  Problems.  Problem  i.  Assume  a  system  of  five  non- 
parallel,  non-concurrent  forces,  and  choose  any  origin  in  their 
plane.  Determine  their  separate  moments,  and  also  the 
moment  of  their  resultant  by  the  method  of  §  62  and  §  63. 

Problem  2.  Given  the  simple  beam  loaded  as  shown  in  Prob- 
lem i,  §  47.  Find  the  moment  at  a  point  12  feet  from  the  left 
end  by  the  method  of  §  64. 

Problem  3.  Given  the  simple  beam  loaded  at  the  same  points 
as  in  Problem  i,  §  47.  The  loads  have  the  same  numerical 
values  as  in  Problem  i,  but  make  the  following  angles  with 
the  axis  of  the  beam  (all  angles  being  measured  counter-clock- 
wise) :  load  at  4-ft.  point,  315°  ;  load  at  center,  270°  ;  load  at 
lO-ft.  point,  240°  ;  load  at  13-ft.  point,  225°.  It  is  required  to 
find  the  bending  moments  in  the  beam  at  points  which  are 
distant  6  ft.,  II  ft.,  and  14  ft.,  respectively,  from  the  left  end  of 
the  beam.  Both  reactions  are  parallel  to  the  resultant  of  all 
the  loads. 


CHAPTER  V. 

CENTER  OF  GRAVITY  OF  AREAS. 

In  designing  static  structures,  it  is  frequently  necessary  to 
deal  with  the  center  of  gravity  of  areas.  The  center  of  grav- 
ity of  some  areas  may  be  found  by  simple  geometrical  con- 
structions; while  for  others,  it  is  convenient  to  divide  the 
figure  into  elementary  areas,  and,  treating  these  areas  as  a 
system  of  parallel  forces,  to  locate  the  point  of  application  of 
the  resultant  of  the  system.  The  point  of  application  of  the 
resultant  of  a  system  of  parallel  forces  acting  at  fixed  points 
is  frequently  referred  to  as  the  center  of  parallel  forces,  but  is 
more  appropriately  called  the  centroid;  and  is  the  point 
through  which  the  line  of  action  of  the  resultant  passes  in 
whatever  direction  the  parallel  forces  are  assumed  to  act. 

Methods  will  be  given  in  this  chapter  for  finding  the  center 
of  gravity  of  some  common  geometrical  areas,  also  of  irregu- 
lar areas. 

66.  Geometrical  Areas.  The  center  of  gravity  of  some  of 
the  most  common  geometrical  areas  will  now  be  located.  The 
proofs  of  the  constructions  will  not  be  given,  as  they  may  be 
found  in  any  treatise  on  plane  geometry ;  but  the  student  should 
be  able  to  supply  the  demonstrations. 

(i)  Parallelogram.  The  center  of  gravity  of  a  parallelo- 
gram is  at  the  point  of  intersection  of  the  two  bisectors  of 
the  opposite  sides  of  the  figure. 
Thus  let  ABCD  (Fig.  31)  be  a  paral- 
lelogram, and  let  ac  and  bd  be  the 
bisectors  of  the  opposite  sides ;  then 
the  center  of  gravity  of  the  figure  is 
at  G,  the  point  of  intersection  of  ac 
and  bd. 

47 


48 


CENTER    OF    GRAVITY    OF    AREAS. 


Chap.   V. 


Triangle. 


C 
FIG.   32. 


FIQ.  33. 


The  center  of  gravity  of  a  triangle  lays  on 
a  line  drawn  from  any  vertex  to  the 
middle  of  the  opposite  side,  and  is 
therefore  at  the  intersection  of  any 
two  such  lines.  The  center  of  gravity 
is  at  a  distance  from  the  vertex  equal 
to  two-thirds  of  the  length  of  any 
bisector.  Thus  let  Ba  (Fig.  32)  be 

one  of  the  bisectors;    then  the  center  of  gravity  G  divides  the 

line  Ba  so  that  BG  is  equal  to  two-thirds  Ba. 

(3)  Quadrilateral.     Let  ABCD  (Fig.  33)  be  a  quadrilateral 
whose  center  of  gravity  is  required.  A 

Draw  BD,  and  let  E  be  its  mid- 
dle point.  Join  E  with  the  points 
A  and  C.  Make  EM  equal  to  one- 
third  EA,  and  EN  equal  to  one- 
third  EC.  Join  the  points  M  and  0 
N,  and  make  MG  =  NH ;  then  G 
is  the  center  of  gravity  of  the  quadrilateral. 

(4)  Circular  Sector.     Let  ABCO   (Fig.  34)   be  a  circular 

sector  whose  center  of  gravity  is  re- 
quired. On  AO  make  Aa  equal  to  one- 
third  AO,  and  describe  the  arc  abc. 
Bisect  the  sector  by  the  line  OB.  Draw 
tangent  to  the  arc  abc  at  b,  and 
make  bd  equal  to  the  length  of  one- 
half  of  the  rectified  arc  abc.  Join  the 
points  O  and  d,  and  from  c  draw 

ce  parallel  and  from  e  draw  eG  perpendicular  to  OB.    Then  G 

is  the  center  of  gravity  of  the  sector. 

(5)  Circular  Segment.     Let  ABC  (Fig.  35)  be  a  circular 
segment   whose    center   of   gravity   is    required.      Draw    the 
extreme  radii  AO  and  CO,  thus  completing  the  sector  ABCO. 
Draw  the  middle  radius  BO.    Locate  the  center  of  gravity  of  the 
triangle  AOC,  and  that  of  the  sector  ABCO  by  the  methods  pre- 
viously explained.     Draw  ab  perpendicular  to  CO,  and  from  g 
and  g',  the  centers  of  gravity  of  the  triangle  ACO  and  the  sector 


CENTROID  OF  PARALLEL  FORCES. 


49 


FIG  35. 


ABCO,  respectively,  draw  gd  and 
g'c  parallel,  in  any  direction.  Make 
g'c  equal  to  ab  and  gd  equal  to  the 
length  of  the  rectified  arc  BC.  Join 
the  points  d  and  c,  and  prolong  the 
line  dc  to  cut  BO  at  G,  which  is  the 
center  of  gravity  of  the  sector. 

67.     Irregular  Areas.  The  cen- 
ter of  gravity  of  an  irregular  area 

may  be  found  by  dividing  the  figure  into  elementary  areas,  re- 
placing these  areas  by  parallel  forces,  and  then  locating  the 
centroid  of  the  resultant  of  the  system  of  forces.  The  centroid 
of  a  system  of  parallel  forces  having  fixed  points  of  applica- 
tion has  been  defined  as  the  point  through  which  the  line  of 
action  of  the  resultant  of  the  system  passes,  in  whatever  direction 
the  forces  are  assumed  to  act;  and  a  method  will  now  be  given 
for  finding  this  centroid. 

68.  Determination  of  the  Centroid  of  Parallel  Forces. 
Let  ab  and  be  (Fig.  36)  be  the  lines  of  action  of  the  two  paral- 
lel forces  AB  and  BC,  and  let  ac  be 
the  line  of  action  of  their  resultant  AC. 
Draw  any  line  MN  perpendicular  to 
the  lines  of  action  of  the  given  forces, 
intersecting  them  at  the  points  M,  O, 
and  N.  Since  AC  is  the  resultant  of 
AB  and  BC,  the  moment  of  AC  about 
any  point  in  their  plane  is  equal  to  the 
sum  of  the  moments  of  AB  and  BC 
about  the  same  point.  If  the  origin  of  moments  is  taken  on 
the  line  of  action  ac,  then  the  moment  of  AC  is  equal  to  zero ; 
and  therefore  —  AB  X  MO  +  BC  X  ON  =o,  or  ABXMO= 
BC  X  ON.  The  latter  equation  shows  that  MN  is  divided  into- 
segments  which  are  inversely  proportional  to  AB  and  BC. 
Any  other  straight  line  cutting  ab  and  be  will  also  be  divided 
into  segments  which  are  inversely  proportional  to  AB  and 
BC.  If  the  two  forces  act  in  opposite  directions,  then  ac  will 
lay  outside  of  ab  and  be;  but  the  above  statement  will  hold 


M 


N 


FIG.  36. 


50  CENTER   OF   GRAVITY    OF   AREAS.  Chap.   V. 

true.  Now  suppose  the  lines  of  action  ab  and  be  to  be  turned 
through  any  angle  about  the  points  M  and  N,  respectively; 
then  the  line  of  action  ac  will  always  pass  through  O.  For,  if 
the  points  M  and  N  remain  fixed,  the  line  of  action  ac  of  the 
resultant  will  always  divide  the  line  MN  into  segments  which 
are  inversely  proportional  to  the  two  forces  AB  and  BC. 
Therefore,  the  point  O  will  always  remain  fixed.  The  point 
O  is  called  the  centroid  of  the  parallel  forces  AB  and  BC  for 
the  fixed  points  M  and  N. 

The  centroid  of  any  number  of  parallel  forces  having  fixed 
points  of  application  may  be  located  by  first  finding  the 
centroid  of  any  two  of  the  forces,  and  then,  taking  the  resultant 
of  these  two  forces  as  acting  at  the  centroid  thus  determined, 
by  finding  the  centroid  of  this  resultant  and  one  of  the  remain- 
ing forces,  etc. 

69.  Graphic  Determination  of  the  Centroid  of  a  System  of 
Parallel  Forces.     Since  the  centroid  must  be  on  the  line  of 
action  of  the  resultant  of  the  given  system  of  forces,  to  locate 
this  centroid  it  is  only  necessary  to  find  the  line  of  action  of 
the  resultant  for  each  of  two  assumed  directions  of  the  forces ; 
then  the  intersection  of  these  two  lines  of  action  is  the  centroid 
of  the  given  system.    The  forces  may  be  assumed  to  be  turned 
through  any  angle ;  but  the  greatest  accuracy  is  secured  if  the 
two  resultants  are  made  to  act  at  right  angles  to  each  other. 

70.  Center  of  Gravity  of  an  Irregular  Area.     The  method 
explained  in  §  69  may  be  used  to  find  the  center  of  gravity  of 
an  irregular  area.    The  figure  may  be  divided  into  areas  whose 
centers  of  gravity  are  known,  and  these  areas  may  then  be 
taken  as  parallel  forces  acting  through  their  respective  centers 
of  gravity.    The  problem  then  resolves  itself  into  finding  the 
centroid  of  a  system  of  parallel  forces  having  fixed  points  of 
application. 

Let  PQRSTUVW  (Fig.  37)  be  the  area  whose  center  of 
gravity  is  required.  Divide  the  figure  into  three  rectangles,  as 
shown,  whose  areas  are  represented  by  the  forces  AB,  BC, 
and  CD.  Then  take  the  point  of  application  of  these  forces 
at  the  centers  of  gravity  of  their  respective  areas,  and  draw 


CENTER   OF    GRAVITY    OF    IRREGULAR   AREAS. 


51 


their  lines  of  action  parallel  to  the  side  PQ.  (The  lines  might 
have  been  drawn  parallel  in  any  direction,  but  are  drawn  as 
indicated  for  convenience).  Construct  the  force  and  the 
funicular  polygons  for  these  forces,  and  locate  the  line  of  action 
of  their  resultant  Rj.  Then  take  the  lines  of  action  of  the 
forces  AB,  BC,  and  CD  parallel  to  the  side  PW,  construct 
their  force  and  funicular  polygons,  and  locate  the  line  of 
action  of  their  resultant  R2.  The  point  of  intersection  of  the 

p  a  A       B       c o 


-r:_aJ.^- 


/  b!c  ^ 


S       R 


\ 

N         \ 

X      \ 


"it--. 


Ul 

i 


w 


.-r;j|'c!d'"v-^ 


i  \ 


FIG.  37.  CENTER  OF  GRAVITY  OF  AN  AREA. 


lines  of  action  of  the  two  resultants  is  the  center  of  gravity  of 
the  area  PQRSTUVW.  For,  the  center  of  gravity  of  the  forces 
which  represent  the  respective  areas  must  be  on  the  line  of 
action  of  every  resultant,  and  is  therefore  at  the  point  of  inter- 
section of  any  two  of  them. 

If  the  given  area  has  an  axis  of  symmetry,  then  its  center 
of  gravity  must  be  on  this  axis;  and  if  it  has  two  such  axes, 
its  center  of  gravity  must  be  at  their  point  of  intersection. 


CHAPTER  VI. 

MOMENT   OF  INERTIA. 

The  moments  of  inertia  of  most  of  the  standard  steel  sections 
used  in  engineering  have  been  computed,  and  may  be  found 
in  tables  published  by  the  manufacturers  of  such  sections.  The 
moment  of  inertia  of  an  irregular  shaped  section  is  often 
required,  and  the  moment  of  inertia  of  such  a  section  may  be 
readily  found  by  graphic  methods.  The  elementary  areas  com- 
posing the  section  may  be  treated  as  parallel  forces,  and  the 
moment  of  inertia  of  these  forces  may  then  be  found  by  means 
of  the  force  and  the  funicular  polygons.  This  chapter  will 
therefore  be  divided  into  two  articles,  as  follows:  Art.  i, 
Moment  of  Inertia  of  Parallel  Forces,  and  Art.  2,  Moment  of 
Inertia  of  Areas. 


ART.  i.     MOMENT  OF  INERTIA  OF  PARALLEL  FORCES. 

71.  Definition.  The  moment  of  inertia  of  a  force  with 
respect  to  any  axis  is  the  product  of  the  magnitude  of  the  force  • 
into  the  square  of  the  distance  of  its  point  of  application  from 
the  given  axis.  The  moment  of  inertia  of  a  system  of  parallel 
forces  with  respect  to  any  axis  is  the  sum  of  the  moments  of 
inertia  of  the  separate  forces  composing  the  system  about  the 
same  axis. 

The  student  should  notice  that  the  moment  of  inertia  as 
defined  above  is  simply  a  mathematical  product  which  fre- 
quently occurs  in  the  solution  of  engineering  problems,  and  in 
no  sense  involves  the  inertia  of  a  body  or  mass.  Since  the 

52 


Art.  i.  CULMANN'S  METHOD.  53 

moment  of  inertia  is  so  frequently  required  in  the  solution  of 
problems,  it  is  desirable  to  determine  this  product  for  all  sec- 
tions likely  to  occur  in  practice. 

72.  Moment  of  Inertia  of  a  System  of  Parallel  Forces. 
There  are  two  methods  in  common  use  for  finding  the  moment 
of  inertia  of  forces,  viz. :   Culmann's,  and  Mohr's.    In  Culmann's 
method,  the  moment  of  inertia  is  determined  by  first  finding 
the  moment  of  the  given  forces,  and  then  the  moment  of  this 
moment,  by  means  of  the  force  and  the  funicular  polygons.    In 
Mohr's  method,  the  moment  of  inertia  is  determined  from  the 
area  of  the  funicular  polygon. 

73.  Culmann's  Method.     Let  the  given  system  of  parallel 
forces  and  the  axis  with  respect  to  which  their  moment  of 
inertia  is  to  be  determined  lay  in  the  same  plane.     The  moment 
of  inertia  of  any  one  of  the  given  forces  may  be  found  as  fol- 
lows:    First  find  the  ^moment  of  the  force  about  the  given 
axis;  then  assume  a  force,  equal  in  magnitude  to  this  moment 
and  acting  in  a  direction   corresponding  to  the  .  sign  of  the 
moment,  to  act  at  the  point  of  application  of  the  original  force, 
and  find  the  moment  of  this  new  force  about  the  given  axis, 
which  is  the  moment  of  inertia  of  the  given  force.  The  algebraic 
sum  of  the  moments  of  inertia  of  all  of  the  forces  is  the  moment 
of  inertia  of  the  given  system.    The  application  of  the  above 
principles  to  a  problem  will  now  be  shown. 

Let  ab,  be,  cd,  and  de  (Fig.  38)  be  the  lines  of  action  of 
the  given  system  of  forces  AB,  BC,  CD,  and'DE  whose  magni- 
tudes and  directions  are  as  shown  in  the  force  polygon.  Also, 
let  the  line  XX,  which  is  drawn  parallel  to  the  lines  of  action 
of  the  given  forces,  be  the  axis  with  respect  to  which  their 
moment  of  inertia  I  is  required. 

Construct  the  force  polygon  ABCDE,  and  the  funicular 
polygon  whose  sides  are  oa,  ob,  oc,  od,  and  oe,  for  the  given 
forces.  Then  the  moment  of  the  force  AB  with  respect  to  the 
axis  XX  is  equal  to  A'B'  X  H  (§62).  Also,  the  moments  of 
the  forces  BC,  CD,  and  DE  are  respectively  equal  to  B'C'  X  H, 
C'D'  X  H,  and  D'E'  X  H.  It  is  seen  from  Fig.  38  that  the 
distinction  between  positive  and  negative  moments  is  given  if 


54 


MOMENT   OF   INERTIA. 


Chap.   VI. 


the  sequence  of  the  letters  is  observed  in  reading  the  intercepts. 
Now  take  these  moments  as  forces,  applied  along  the  lines  of 
action  of  the  original  forces.  Thus,  let  the  force  represented 
by  A'B'  X  H  be  applied  along  ab,  let  the  force  represented  by 
B'C'  X  H  be  applied  along  be,  etc.  The  line  A'B'C'D'E'  may 
be  taken  as  the  force  polygon  for  the  second  system  of  forces ; 
but  it  must  be  borne  in  mind  that  each  force  in  this  polygon 
must  be  multiplied  by  H  to  give  its  true  magnitude.  Assume 
any  pole  O',  and  construct  a  new  funicular  polygon  whose 
sides  are  o'a',  o'b',  o'c',  o'd',  and  o'e'.  Since  the  moment  of 
inertia  of  the  system  of  forces  is  required,  and  since  the  sum 
of  the  separate  intercepts  is  equal  to  the  intercept  between  the 
extreme  strings,  it  is  only  necessary  to  prolong  the  extreme 
strings  o'a'  and  o'e'  until  they  intersect  the  axis  XX  at  the 

X 


H' 


r^--f--^^ 


A 

B^ 

CV 

^ 
>» 
>x 

[>>.x< 

£^ 

D' 
L' 

-"^      / 
/ 
/ 

s 
/ 
/ 

's 

FIG.  38.     MOMENT  OF  INERTIA — CULMANN'S  METHOD. 

points  M  and  N,  respectively.  The  moment  of  this  second 
system  of  forces  is  then  equal  to  MN  X  H  X  H',  which  is  the 
moment  of  inertia  of  the  given  system.  For,  A'E'  X  H  is  the 
moment  of  the  original  system  of  forces  AB,  BC,  CD,  and  DE 
(§  63),  and  is  equal  to  the  summation  of  the  products  of  each 
force  into  its  distance  from  the  axis  XX ;  also  MN  X  H  X  H' 


Art.  1. 


MOHR  S  METHOD. 


55 


is  the  moment  of  this  moment,  and  is  equal  to  the  summation 
of  the  products  of  each  force  into  the  square  of  its  distance 
from  the  axis  XX,  which  by  definition  is  the  moment  of 
inertia  I  of  the  given  system  of  forces. 

It  should  be  noted  that  H  is  a  line  in  the  force  diagram, 
therefore  H  represents  a  force  and  is  measured  to  the  same 
scale  as  the  given  forces;  while  H'  and  MN  are  lines  in  the 
space  diagram,  represent  distances,  and  are  measured  to  the 
same  scale  as  the  distances  in  the  space  diagram.  In  choosing 
H  and  H',  they  should  be  taken  of  such  units  of  length  that  the 
numerical  force  which  H  represents  and  the  distance  which 
H'  represents  may  be  easily  multiplied  together. 

74.  Mohr's  Method.  Let  ab,  be,  cd,  and  de  (Fig.  39)  be 
the  lines  of  action  of  the  given  forces  AB,  BC,  CD,  and  DE, 
whose  magnitudes  and  directions  are  as  shown  in  the  force 
polygon.  It  is  required  to  find  the  moment  of  inertia  of  the 
system  about  the  axis  XX,  which  axis  is  parallel  to  the  lines  of 
action  of  the  given  forces. 


A 

x» 

X 

X 

B' 

X 

C< 

""""VJo 

D' 

/ 

/ 

FIG.  39.     MOMENT  OF  INERTIA — MOHR'S  METHOD. 

Construct  the  force  polygon  ABCDE,  and  with  the  pole  O 
and  the  pole  distance  H,  draw  the  funicular  polygon  whose 
strings  are  oa,  ob,  oc,  od,  and  oe.  Prolong  these  strings  until 
they  intersect  the  axis  XX  at  the  points  A',  B',  C',  D',  and 
E',  respectively.  Then  the  moment  of  AB  about  XX  is  equal 


56  MOMENT   OF  INERTIA.  Chap.   VI. 

to  the  intercept  A'B'  multiplied  by  the  pole  distance  H ;  and 
the  moment  of  inertia  of  AB  about  the  same  axis  is  equal  to 
the  moment  of  this  moment,  which  is  equal  to  A'B'  X  H  X  rn. 
But  A'B'  X  rn  equals  twice  the  area  of  the  triangle  whose 
base  is  A'B'  and  whose  vertex  is  on  ab ;  and  therefore  the 
moment  of  inertia  of  AB  about  the  axis  XX  is  equal  to  the 
area  of  this  triangle  multiplied  by  2H.  In  like  manner,  it  may 
be  shown  that  the  moment  of  inertia  of  BC  about  the  axis  XX 
is  equal  to  the  area  of  the  triangle  whose  base  is  B'C'  and 
whose  vertex  is  on  be,  multiplied  by  2H ;  also  that  the  moment 
of  inertia  of  CD  is  equal  -to  the  area  of  the  triangle  whose  base 
is  C'D'  and  whose  vertex  is  on  cd,  multiplied  by  2H ;  and 
further  that  the  moment  of  inertia  of  DE  is  equal  to  the  area 
of  the  triangle  whose  base  is  D'E'  and  whose  vertex  is  on 
de,  multiplied  by  2H.  Adding  the  moments  of  inertia  of  these 
forces,  it  is  seen  that  the  moment  of  inertia  of  the  given  system 
is  equal  to  the  area  of  the  funicular  polygon  multiplied  by  twice 
the  pole  distance,  i.e.,  I  is  equal  to  the  area  of  the  polygon  whose 
sides  are  oa,  ob,  oc,  od,  oe,  and  E'A',  multiplied  by  2H. 

75.  Relation  Between  Moments  of  Inertia  About  Parallel 
Axes.  A  graphic  proof  will  now  be  given  for  the  proposition 
that  the  moment  of  inertia  I  of  a  system  of  parallel  forces 
about  any  axis  parallel  to  the  forces  is  equal  to  the  moment  of 
inertia  Icg  about  an  axis  through  their  centroid  plus  the  moment 
of  inertia  Ir  of  the  resultant  of  the  system  about  the  given  axis. 

The  moment  of  the  resultant  R  of  the  system  of  forces 
shown  in  Fig.  39  about  the  axis  XX  is  equal  to  A'E'  X  H ; 
and  the  moment  of  inertia  Ir  of  R  .about  XX  is  equal  to  the 
moment  of  this  moment,  which  is  equal  to  A'E'  X  H  X  h 
(where  h  is  the  perpendicular  distance  between  XX  and  the 
line  of  action  of  R).  But  A'E'  X  h  is  equal  to  twice  the  area 
of  the  triangle  whose  base  is  A'E'  and  whose  vertex  is  on  the 
line  of  action  of  the  resultant  R.  Therefore,  the  moment  of 
inertia  If  of  R  about  XX  is  equal  to  the  area  of  the  funicular 
triangle  whose  base  is  A'E'  and  whose  vertex  is  on  the  line 
of  action  of  R,  multiplied  by  2H.  In  like  manner,  if  the  axis 
is  taken  to  coincide  with  the  line  of  action  of  R,  i.e.,  through 


Art.  1.  RADIUS  OF  GYRATION.  57 

the  centroid  of  the  system,  it  may  be  shown  'that  the  moment 
of  inertia  Ic.*.  is  equal  to  the  area  of  the  funicular  polygon 
whose  sides  are  oa,  ob,  oc,  od,  and  oe,  multiplied  by  2H. 
Now  the  area  of  the  triangle  plus  the  area  of  the  polygon  is 
equal  to  the  total  area  of  the  figure  whose  sides  are  oa,  ob,  oc, 
od,  oe,  and  E'A'.  But  the  total  area  of  this  figure  multiplied 
by  2H  is  equal  to  the  moment  of  inertia  I  of  the  given  system 
of  forces  about  the  axis  XX;  therefore,  I  =  Ic.e.  +  Ir  . 

It  will  be  shown  that  the  moment  of  inertia  of  a  system 
of  forces  about  any  axis  is  equal  to  the  moment  of  inertia 
of  the  given  system  about  a  parallel  axis  through  its  centroid 
plus  the  product  of  the  magnitude  of  the  resultant  of 
the  system  into  the  square  of  the  distance  between  the  two 
axes.  For,  by  definition,  the  moment  of  inertia  Ir  of  the 
resultant  of  the  system  about  the  axis  XX  is  equal  to  Rh2 
(where  h  is  the  perpendicular  distance  from  the  line  of  action 
of  R  to  the  axis  XX).  Substituting  this  value  of  It  in  the 
equation  I  =  Ic.e.  +Ir  gives  I  =  Ic.e.  -|~  Rh2,  which  proves 
the  proposition. 

By  taking  the  axis  through  the  centroid  of  the  system  and 
then  moving  it,  first  to  one  side,  then  to  the  other  side  of  this 
centroid,  it  is  readily  seen  (Fig.  39)  that  any  movement  of  the 
axis  out  of  the  centroid  increases  the  total  area  included 
between  the  extreme  strings  meeting  on  the  resultant.  There- 
fore, the  moment  of  inertia  of  a  given  system  of  parallel  forces 
is  a  minimum  about  an  axis  through  the  centroid  of  the 
system. 

76.  Radius  of  Gyration.  The  radius  of  gyration  of  a  sys- 
tem of  parallel  forces  about  any  axis  is  the  distance  from  the 
axis  to  a  point  through  which  the  resultant  of  the  system 
must  act  in  order  that  the  moment  of  inertia  may  be  the  same 
as  that  of  the  given  system.  An  equation  will  now  be  derived 
for  the  radius  of  gyration  in  terms  of  the  moment  of  inertia 
and  the  resultant  of  the  given  system.  Let  R,  the  resultant 
of  the  system,  be  substituted  for  the  given  forces,  and  let  this 
resultant  act  at  such  a  distance  r  (which  by  definition  is  the 
radius  of  gyration  of  the  system)  from  the  axis  that  its 


58  MOMENT   OF  INERTIA.  Chap.   VI. 

moment  of  inertia  I±  remains  the  same  as  that  of  the  original 
system  of  forces,  I.    Then  Ij  =  I  =  Rr2, 


or  r2  =  — — ,  which  gives  r  =  .  /  — 
R  \   R 


77.  Graphic  Determination  of  the  Radius  of  Gyration. 
The  moment  of  inertia  of  the  system  of  forces  shown  in  Fig.  38 
is  equal  to  H  X  H'  X  MN.  If  r  is  the  radius  of  gyration  of 

the  given  system,  then   (§  76),  r2=^-=  H  X  H/  X  MN  . 

R  R 

Now  if  H  (Fig.  38)  is  taken  equal  to  AE  =  R,  then  the  preced- 
ing   equation    becomes    r2  =  H'XMN,    which     gives     r  = 


'  X  MN.     The  length  of  this  radius  of  gyration  may  be  found, 
graphically,    as    follows  :      Draw    AB  =  MN,    and    BC  =  H'. 

With  AC  as  a  diameter,  construct  the 
semi-circle  ADC;  and  from  B,  draw 
the  line  BD  perpendicular  to  AC,  in- 
tersecting the  semi-circle  at  D.  Then 
from  geometry,  AB  :  BD  :  :  BD  :  BC, 
or  AB  X  BC  —  BD2.  Substituting  the 
FIG-  40-  values  of  AB  '  and  BC,  we  have 


MN  X  H'  =  BD  ,  or  BD  =  \/H/  X  MN,  which  has  been  shown 
equal  to  the  radius  of  gyration  r. 

In  the  above  construction,  H  has  been  taken  equal  to 
AE  =  R.  If  H  had  been  taken  equal  to  nR,  then  AB  (Fig.  40) 
should  have  been  taken  equal  to  nMN,  or  BC  equal  to  nH'. 

ART.  2.     MOMENT  OF  INERTIA  OF  AREAS. 

78.  Moment  of  Inertia  of  an  Area.  The  moment  of  inertia 
of  an  area  about  any  axis  is  equal  to  the  summation  of  the 
products  of  the  differential  areas  composing  the  area  into  the 
squares  of  the  distances  of  these  differential  areas  from  the 
axis.  The  moment  of  inertia  of  an  area  may  be  found  by 
dividing  the  given  area  into  elementary  areas,  treating  these 
elements  as  parallel  forces,  and  finding  the  moment  of  inertia 


Art. 


APPROXIMATE    METHOD  -  MOMENT   OF   INERTIA. 


59 


of  the  forces.  There  are  two  somewhat  similar  graphic 
methods  for  finding  the  moment  of  inertia  of  an  area,  one  of 
which  gives  an  approximate  value,  and  the  other  an  accurate 
value. 

79.  Approximate  Method  for  Finding  the  Moment  of 
Inertia  of  an  Area.  An  approximate  value  for  the  moment  of 
inertia  of  an  area  about  any  axis  may  be  obtained  as  follows : 
Divide  the  given  area  into  small  elements  by  lines  drawn 
parallel  to  the  axis,  and  let  these  elementary  areas  be  replaced 
by  forces  numerically  equal  to  them,  acting  at  their  respective 
centers  of  gravity.  Then  find  the  moment  of  inertia  of  this 
system  of  forces,  which  will  be  approximately  equal  to  that  of 
the  given  area.  The  smaller  these  elementary  areas  are  taken, 
the  more  nearly  will  the  moment  of  inertia  of  the  parallel 
forces  which  represent  them  approximate  that  of  the  given 
area ;  and  if  they  could  be  taken  as  infinitesimal  in  size,  their 
moment  of  inertia  would  be  the  true  value  for  the  moment  of 
inertia  of  the  area. 

The  application  of  the  above  principles  to  the  determina- 
tion of  the  moment  of  inertia  of  an  area  will  now  be  shown. 


FIG.  41. 


60  MOMENT  OF  INERTIA.  Chap.   VI. 

Let  the  area  shown  in  Fig.  41  be  the  area  whose  moment  of 
inertia  about  the  axis  XX  is  required. 

Divide  the  section  into  four  rectangular  areas  as  shown, 
and  assume  the  forces  AB,  BC,  CD,  and  DE,  numerically 
equal  to  these  areas  and  parallel  to  the  axis  XX,  to  act  at  the 
centers  of  gravity  of  the  respective  areas.  Construct  the  force 
polygon  ABCDE  for  these  forces,  and  with  the  pole  distance 
H  (in  this  case  H  is  taken  equal  to  AE)  and  the  pole  O,  draw 
the  funicular  polygon  corresponding  to  this  pole.  Prolong 
the  strings  of  the  funicular  polygon  until  they  intersect  the 
axis  at  the  points  A',  B',  C',  D',  and  E',  and  with  the  pole 
distance  H'  and  the  pole  O',  draw  the  second  funicular  polygon, 
using  the  same  lines  of  action  as  before.  Prolong  the  first  and 
last  strings  of  this  funicular  polygon  until  they  intersect  the 
axis  XX  at  the  points  M  and  N.  Then  H  X  H'  X  MN  is  equal 
to  the  moment  of  inertia  of  the  forces  AB,  BC,  CD,  and  DE, 
(§  73) >  and  is  approximately  equal  to  the  moment  of  inertia  of 
the  given1  area. 

A  more  nearly  correct  value  for  the  moment  of  inertia  of 
the  area  might  have  been  obtained  if  the  area  had  been  divided 
into  smaller  strips  by  lines  drawn  parallel  to  XX ;  as  the 
distance  of  each  elementary  area  from  the  axis  would  then 
have  been  more  nearly  equal  to  the  distance  of  its  center  of 
gravity  from  the  axis. 

80.  Radius  of  Gyration  of  an  Area.  The  radius  of  gyra- 
tion of  an  area  about  any  axis  is  the  distance  from  the  axis  to 
a  point  at  which  if  all  the  area  was  concentrated  the  moment 
of  inertia  would  be  the  same  as  that  of  the  given  area. 

An  approximate  value  for  the  radius  of  gyration  of  the  area 
shown  in  Fig.  41  may  be  obtained  by  applying  the  method 
explained  in  §  77.  Thus,  lay  off  BC  (Fig.  41,  a)  equal  to  MN, 
and  AB  equal  to  H'.  Then  on  AC  as  a  diameter,  construct  a 
semi-circle,  and  from  B  draw  the  line  BD  perpendicular  to  AC, 
cutting  the  semi-circle  at  the  point  D.  Then  BD  is  approxi- 
mately equal  to  the  radius  of  gyration  of  the  given  area.  It 
should  be  remembered  that  the  pole  distance  H  (Fig.  41)  was 
taken  equal  to  AE.  If  H  had  been  taken  equal  to  n  X  AE, 


Art.  2.  ACCURATE    METHOD  —  MOMENT   OF   INERTIA.  61 

then  AB  should  have  been  taken  equal  to  nH',  or  BC  equal 
to  nMN. 

81.  Accurate  Method  for  Finding  the  Moment  of  Inertia 
of  an  Area.  An  accurate  value  for  the  moment  of  inertia  of  an 
area  about  any  axis  may  be  obtained  if  it  is  possible  to  divide 
the  area  into  elementary  areas  whose  radii  of  gyration  about 
axes  through  their  centers  of  gravity  and  parallel  to  the  given 
axis  are  known.  For,  if  the  radius  of  gyration  of  each  elemen- 
tary area  about  an  axis  through  its  center  of  gravity  and  also 
the  distance  of  its  center  of  gravity  from  the  given  axis  are 
known,  then  its  radius  of  gyration  about  the  given  axis  may 
be  determined.  Now,  if  forces  numerically  equal  to  these  areas 
and  acting  parallel  to  the  axis  at  distances  equal  to  their 
respective  radii  of  gyration  about  the  given  axis  are  sub- 
stituted for  these  areas,  and  the  moment  of  inertia  of  these 
forces  is  determined,  the  result  will  be  the  moment  of  inertia 
of  the  given  area.  For,  by  definition,  the  radius  of  gyration  of 
an  area  is  the  distance  from  the  axis  to  a  point  at  which  the 
entire  area  must  be  concentrated  "in  order  that  the  moment  of 
inertia  may  remain  the  same  as  about  the  given  axis. 

The  application  of  the  above  method  to  the  determination 
of  the  moment  of  inertia  of  an  area  will  now  be  shown  by  a 
problem.  Let  the  area  shown  in  Fig.  42  be  the  area  whose 
moment  of  inertia  about  XX  is  required,  and  let  AB,  BC,  CD, 
and  DE  represent  in  magnitude  the  elementary  areas  into 
which  the  section  is  divided. 

The  lines  of  action  of  these  forces  may  be  found  as  follows : 
Draw  PR  (Fig.  42)  perpendicular  to  XX  and  equal  to  the 
distance  from  XX  to  the  center  of  gravity  of  the  elementary 
area  shown ;  also,  at  the  point  R,  draw  QR  perpendicular  to 
PR  and  equal  to  the  radius  of  gyration  of  this  area  about  an 
axis  through  its  center  of  gravity  and  parallel  to  XX.  Connect 
the  points  P  and  Q,  and  with  the  line  PQ  as  a  radius,  draw 
the  arc  QS.  Then  PS  =  PQ  is  the  radius  of  gyration  of  the 
elementary  area  about  XX,  and  is  the  distance  from  the  axis 
to  the  point  of  application  of  the  force  AB.  For,  the  moment 
of  inertia  of  the  elementary  area  about  an  axis  through  its 


62 


MOMENT   OF   INERTIA. 


Chap.   VI. 


center  of  gravity  and  parallel  to  XX  is  equal  to  AB  X  QR2 
(from  the  definition  of  the  radius  of  gyration),  and  the  moment 
of  inertia  of  this  area  about  the  axis  XX  is  equal  to 


AB  X  PS2  (§  75).     In  like  manner,  the  lines  of  action  of  the 
forces  BC,  CD,  and  DE,  which  represent  the  other  elementary 

yv  ^f 


\     /   :|   !F 

\  /  U4 


'(  i^k'Tx'; )' 
•-  .--'  H-'  -\  /  \ 


Fia.  42. 


areas,  may  be  found.  Now  apply  the  method  explained  in  §  73 
and  §  79,  and  find  the. moment  of  inertia  of  these  forces  (see 
Fig.  42).  Then  the  moment  of  inertia  of  these  forces,  which 
is  equal  to  the  moment  of  inertia  of  the  given  area,  is  equal  to 
H  X  H'  X  MN. 

The  correct  value  for  the  radius  of  gyration  of  the  area 
shown  in  Fig.  42  is  given  by  the  line  BD  (Fig.  42,  a).  The 
construction  for  finding  this  radius  of  gyration  is  similar  to 
that  used  for  Fig.  41,  a,  which  is  explained  in  §  80,  except  that 
the  values  for  Hr  and  MN,  which  have  been  found  by  the  con- 
struction of  Fig.  42,  are  used  in  Fig.  42,  a. 


Art. 


MOMENTS  OF  INERTIA  ABOUT  PARALLEL  AXES. 


63 


82.  The  table  shown  in  Fig.  42,  b  gives  the  algebraic  for- 
mulae for  finding  the  moments  of  inertia  and  radii  of  gyration 
of  several  common  sections. 


SECTION 

MOMENT  OF  INERTIA 

RADIUS  OF  GYRATION 

d.4 

d 
VT2 

-e- 

can 

_.^_. 

\Z 

bds-b,d,3 
12 

vff 

-/  bd^bd? 

1 

Vl2(bd-b,d.) 

l-dj 

0.049  d4 

i 

~fi~ 

0.049(d^d4) 

VSS 

FIG.  42,  b. 

83.  Relation  Between  the  Moments  of  Inertia  of  an  Area 
About  Parallel  Axes.     If  the  areas  are  used  instead  of  the 
forces  which  represent  them,  then  the  relation  shown  in  §  75 
for  the  moment  of  inertia  of  a  force  about  any  axis,  in  terms  of 
its  moment  of  inertia  about  a  parallel  axis  through  its  center 
of  gravity,  becomes,  the  moment  of  inertia  of  an  area  about 
any  axis  is  equal  to  its  moment  of  inertia  about  a  parallel  axis 
through  its  center  of  gravity  plus  the  product  of  the  area  into 
the  square  of  the  distance  between  the  two  axes.    The  applica- 
tion of  this  relation  is  of  great  importance  in  designing  sec- 
tions. 

84.  Moment  of  Inertia  of  an  Area  Determined  from  the 
Area  of  the   Funicular  Polygon    (Mohr's   Method).     In  the 


64  MOMENT  OF  INERTIA.  Chap.  VI. 

examples  that  have  been  given  for  finding  the  moment  of 
inertia  of  an  area,  only  Culmann's  method  has  been  used,  but 
Mohr's  method  might  have  been  employed  instead.  If  the 
lines  of  action  of  the  forces  had  been  taken  as  acting  at  the 
centers  of  gravity  of  the  elementary  areas,  an  approximate 
value  for  the  moment  of  inertia  would  have  been  found ;  and  if 
each  force  had  been  taken  as  acting  at  a  distance  from  the 
given  axis  equal  to  the  radius  of  gyration  of  the  elementary 
area  about  the  given  axis,  then  an  accurate  value  would  have 
been  found  from  the  area  of  the  funicular  polygon. 


PART  II. 

FRAMED  STRUCTURES-ROOF  TRUSSED 

CHAPTER  VII. 
DEFINITIONS. 

85.  Framed  Structure.     A  framed  structure  is  a  structure 
composed  of  a  series  of  straight  members  fastened  together  at 
their  ends  in  such  a  manner  as  to  make  the  entire  frame  act  as 
a  rigid  body. 

The  only  geometrical  figure  which  is  incapable  of  any 
change  of  shape  without  a  change  in  the  length  of  its  sides  is 
the  triangle;  and  it  therefore  follows  that  the  triangle  is  the 
basis  of  the  arrangement  of  the  members  in  a  framed  structure. 

86.  Types  of  Framed  Structures.     Framed  structures  may 
be   divided   into   the   three   following   classes:     (a)    complete 
framed  structures,  (b)  incomplete  framed  structures,  and  (c) 
redundant  framed  structures. 

(a)  Complete  Framed  Structure.  A  complete  framed 
structure  is  one  which  is  made  up  of  the  minimum  number  of 
members  required  to  form  the  structure  wholly  of  triangles. 
This  is  the  type  which  is  usu- 
ally employed,  and  which  will 
receive  the  most  attention  in 
this  work.  A  simple  form  of  a 
complete  framed  structure  is 

Shown  in  Fig.  43.  FIG.   43.     COMPLETE. 

65 


66 


DEFINITIONS. 


Chap.  VII. 


(b)     Incomplete  Framed  Structure.     An  incomplete  framed 
structure  is  one  which  is  not  wholly  composed  of  triangles. 

A  simple  form  of  an 
incomplete  framed 
structure  is  shown 
in  Fig.  44.  Such  a 
structure  is  stable 
only  under  symmet- 


FI'G.  44.     INCOMPLETE. 


rical  or  specially  ar- 
ranged loads. 

(c)  Redundant  Framed  Structure.  A  redundant  framed 
structure  is  one  which  contains  a  greater  number  of  members 
than  is  required  to  form  the  structure  wholly  of  triangles.  If 
the  second  diagonal  is  added  to  a  quadrilateral,  then  the  added 
diagonal  is  a  redund- 
ant member;  but  if 
the  member  is  capa- 
ble of  resisting  only 
one  kind  of  stress, 
then  the  redundancy 

is  Only  apparent.  Fig.  FlQ-  44'  a-     REDUNDANT. 

44,  a  shows  a  redundant  structure;  but  if  the  diagonals  are 
made  of  rods  and  can  therefore  take  tension  only,  then  the 
redundancy  is  only  apparent;  as  only  one  diagonal  will  act 
at  a  time. 

87.     Roof  Truss.     A  simple  roof  truss  is  a  framed  struc- 
ture whose  plane  is  vertical  and  which  is  supported  at  its 


Web  Members 


<- Lower  Chord  3 
Soon 


PIG.  45.     ROOF  TRUSS. 


TYPES  OF  ROOF  TRUSSES. 


67 


ends.  The  ends  of  the  truss  may  be  supported  upon  side 
walls,  or  upon  columns.  A  common  form  of  a  simple  roof 
truss  is  shown  in  Fig.  45. 

Span.     The  span  of  a  truss  is  the  distance  between  the  end 
joints,  or  the  centers  of  the  supports,  of  the  truss. 


8  Panels 


16  Panels 
(d) 


With  Ventilator 
(f) 


Circular  Chord 
(h) 


Howe 


10  Panels 
(b) 


12  Panels 

(C) 


20  Panels 

(6) 


Cambered 


FINK  TRUSSES 


Quadrangular 


Pratt 
(k) 


Saw  Tooth 


Cantilever 

(I)  (m) 

FIG.   46.     TYPES   OF   ROOF  TRUSSES. 


68  DEFINITIONS.  Chap.   VII. 

Rise.  The  rise  is  the  distance  from  the  apex,  or  highest 
point  of  the  truss,  to  the  line  joining  the  points  of  support  of  the 
truss. 

Pitch.  The  pitch  is  the  ratio  of  the  rise  of  the  truss  to  its 
span. 

Upper  and  Lower  Chords.  The  upper  chord  consists  of  the 
upper  line  of  members  of  the  truss,  and  extends  from  one  support 
to  the  other,  through  the  apex.  The  lower  chord  consists  of  the 
lower  line  of  members,  and  extends  from  one  support  to  the  other. 

Web  Members.  The  web  members  connect  the  joints  of  the 
upper  chord  with  those  of  the  lower  chord.  A  web  member 
which  is  subject  to  compression  is  called  a  strut,  and  one  which 
is  subject  to  tension  is  called  a  tie. 

Pin-connected  and  Riveted  Trusses.  A  pin-connected  truss  is 
one  in  which  the  members  are  connected  with  each  other  by  means 
of  pins.  A  riveted  truss  is  one  in  which  the  members  are  fastened 
together  by  means  of  plates  and  rivets.  The  latter  type  is  the 
more  common,  while  the  former  is  used  for  long  spans. 

88.  Types  of  Roof  Trusses.  Several  types  of  roof 
trusses  are  shown  in  Fig.  46. 

When  a  building  is  to  be  designed,  the  conditions  govern- 
ing the  design  should  be  carefully  studied  before  deciding 
upon  the  type  of  roof  truss  to  be  used ;  as  a  truss  which  is 
economical  for  one  building  may  not  prove  so  for  another. 

The  Fink  truss  is  very  commonly  used,  and  is  well 
adapted  to  different  spans ;  as  the  number  of  panels  may  be  easily 
increased. 

The  Quadrangular  truss  shown  in  Fig.  46,  i  is  well  adapted 
to  the  use  of  counterbracing. 

The  Howe  and  Pratt  trusses  shown  in  Fig.  46,  j  and  Fig. 
46,  k,  respectively,  are  common  type,  and  may  be  used  with 
a  small  rise. 

As  a  rule,  trusses  with  long  compression  members  are  not 
economical. 


CHAPTER  VIII. 

LOADS. 

The  loads  that  must  be  considered  in  the  discussion  of  a 
roof  truss  may  be  classed  as  follows:  (i)  dead  loads;  (2) 
snow  loads;  and  (3)  wind  loads.  A  discussion  of  these  loads 
will  be  given  in  the  three  following  articles. 

Instead  of  considering  separately  the  dead,  snow,  and 
wind  loads,  an  equivalent  vertical  load  is  sometimes  taken. 
This  method  is  efficient  in  some  cases  if  intelligently  used, 
but  should  not  be  used  by  the  beginner. 

ART.  i.     DEAD  LOAD. 

A  short  description  of  the  common  type  of  roof  construc- 
tion, together  with  the  terms  employed,  will  now  be  given, 
preliminary  to  the  discussion  of  dead  loads. 

89.  Construction  of  a  Roof.  A  roof  includes  the  covering 
and  the  framework.  There  are  a  number  of  materials  used 
for  roof  coverings,  among  the  more  common  being  slate,  tiles, 
tar  and  gravel,  tin,  and  corrugated  steel.  Sheathing  is  com- 
monly used  in  connection  with  roof  coverings,  but  it  is  often 
omitted  when  corrugated  steel  is  used. 

The  covering  is  usually  supported  by  members  called 
jack-rafters,  which  in  turn  are  supported  by  other  members 
called  purlins.  The  purlins  run  longitudinally  with  the  build- 
ing and  are  connected  to  the  trusses,  generally  at  panel  points. 
The  jack-rafters  are  usually  made  of  wood,  while  the  purlins 
may  be  either  of  wood,  or  of  steel.  If  the  distance  between 
the  trusses  is  great,  the  purlins  may  be  trussed. 

69 


70  LOADS.  Chap.   VIII. 

A  system  of  sway  bracing  is  generally  used  to  give  rigidity 
to  the  structure.  This  bracing  may  be  in  the  plane  of  the 
upper  chord,  in  the  plane  of  the  lower  chord,  or  in  both  planes. 
Sometimes  the  sway  bracing  is  made  continuous  throughout 
the  entire  length  of  the  building,  and  at  other  times  the 
trusses  are  connected  in  pairs,  depending  upon  the  rigidity 
required. 

The  trusses  may  be  supported  upon  masonry  walls,  upon 
masonry  piers,  or  upon  columns.  When  the  trusses  rest  upon 
masonry  and  the  span  is  short,  the  expansion  and  contraction 
of  the  trusses  are  provided  for  by  a  planed  base  plate ;  but  for 
long  spans,  rollers  are  usually  employed. 

90.  Dead  Load.  The  dead  load  includes  the  weights  of 
the  following  items:  (i)  roof  covering;  (2)  purlins,  rafters, 
and  bracing;  (3)  roof  trusses;  and  (4)  permanent  loads  sup- 
ported by  the  trusses. 

1 i )  Roof  Covering.     The  weight  of  the  roof  covering  varies 
greatly,   depending  upon   the   materials   employed;    but   it   may 
be  closely  estimated  if  the  materials  used  in  its  construction 
are  known.    The  approximate  weights  of  some  of  the  common 
roof  coverings  are  given  in  the  following  table : 

Slate,  without  sheathing 8  to  10  Ibs.  per  sq.  ft. 

Tiles,  flat 15  to  20  Ibs.  per  sq.  ft. 

Tiles,   corrugated 8  to  10  Ibs.  per  sq.  ft. 

Tar  and  gravel,  without  sheathing. .  8  to  10  Ibs.  per  sq.  ft. 

Tin,  without  sheathing i  to  1.5  Ibs.  per  sq.  ft. 

Wooden  shingles,  without  sheathing  2  to  3     Ibs.  per  sq.  ft. 

Corrugated  steel,  without  sheathing  i  to  3     Ibs.  per  sq.  ft. 

Sheathing,  I  inch  thick 3  to  4  Ibs.  per  sq.  ft. 

(2)  Purlins,  Rafters,  and  Bracing.     Wooden  purlins  will 
weigh  from  1.5  to  3  pounds,  and  steel  purlins  from  1.5  to  4  pounds 
per  square  foot  of  roof  surface. 

Rafters  will  weigh  from  1.5  to  3  pounds  per  square  foot  of 
roof  surface. 

The  weight  of  the  bracing  is  a  variable  quantity,  depending 
upon  the  rigidity  required.    If  bracing  is  used  in  the  planes  of 


^rt-  1-  DEAD    LOAD.  71 

both  the  upper  and  lower  chords,  its  weight  will  be  from  0.5  to  I 
pound  per  square  foot  of  roof  surface. 

(3)  Roof  Trusses.  The  roof  trusses  may  be  either  of  wood 
or  of  steel.  Steel  trusses  are  now  commonly  used  for  spans  of 
considerable  length,  although  wooden  trusses  are  still  used  for 
short  spans ;  but  the  rapidly  increasing  cost  of  wood,  and  the 
difficulty  of  framing  the  joints  so  as  to  develop ,  the  entire 
strength  of  the  members,  has  led  to  a  general  use  of  steel  for 
trusses. 

The  weights  of  wooden  trusses  are  given  by  the  following 
formula,  taken  from  Merriman's  "Roofs  and  Bridges,"  which 
is  based  upon  data  given  in  Kicker's  "Construction  of  Trussed 
Roofs." 

W=-iAL(i+^0L),  (i) 

where       W  =  the  total  weight  of  one  truss  in  pounds, 

A  =  the  distance  between  adjacent  trusses  in  feet, 
L  =  the  span  of  the  truss  in  feet. 

The  weights  of  steel  roof  trusses  vary  with  the  span,  the 
pitch,  the  distance  between  trusses,  and  the  capacity  or  load 
carried  by  the  trusses.  The  following  formula  is  given  in 
Ketchum's  "Steel  Mill  Buildings,"  and  is  based  upon  the  actual 
shipping  weights  of  trusses : 

W=45-AL(I  +  7VA}'  (2) 

where     W  =  weight  of  the  steel  roof  truss  in  pounds, 

p  =  capacity  of  the  truss  in  pounds  per  square  foot 

of  horizontal  projection  of  the  roof, 
A  =  distance,  center  to  center  of  trusses,  in  feet, 
L  =  span  of  truss  in  feet. 
These  trusses  were  designed  for  a  tensile  stress  of  15000  Ibs. 

per  sq.  in.,  and  a  compressive  stress  of  15000  —  55  —  Ibs.  per 

sq.  in. ;  where  1  =  length,  and  r  =  radius  of  gyration  of  the 
member,  both  in  inches.  The  minimum  sized  angle  used  was 


72  LOADS.  Chap.   VIII. 

2"  X  2"  X  %",  and  the  minimum  thickness  of  plate  was  one- 
fourth  inch. 

Dividing  equation  (2)  by  AL,  we  have 

w:=  —   (i+— k-),  (3) 

45  5VA'1 

where  w  is  the  weight  in  Ibs.  per  sq.  ft.  of  the  horizontal  pro- 
jection of  the  roof. 

Short  span  simple  roof  trusses  may  weigh  somewhat  less 
than  the  values  given  by  the  above  formulae,  especially  if  the 
minimum  thickness  and  minimum  size  of  angles  are  used,  but 
such  trusses  are  too  light  to  give  good  service. 

(4)  Permanent  Loads  Supported  by  the  Trusses.  It  is  im- 
possible to  give  figures  which  will  be  of  much  value  for  the 
weights  of  permanent  loads  supported  by  the  trusses.  If  the 
roof  truss  supports  a  plastered  ceiling,  the  weight  of  the  ceil- 
ing may  be  taken  at  about  10  Ibs.  per  sq.  ft.  If  other  loads 
are  supported  by  the  truss,  they  should  be  carefully  consid- 
ered in  estimating  the  weight  of  the  truss. 


ART.  2.     SNOW  LOAD. 

91.  Snow  Load.  The  amount  of  the  snow  load  to  be  con- 
sidered varies  greatly  for  different  localities.  The  weight  of 
the  snow  which  must  be  taken  into  account  is  a  function  of 
both  the  latitude  and  the  humidity,  and  those  factors  should 
be  carefully  considered  in  determining  the  weight  to  be  used 
for  any  particular  locality.  The  maximum  wind  load  and  the 
maximum  snow  load  will  probably  never  occur  at  the  same 
time ;  and  if  the  maximum  wind  is  taken,  then  a  smaller  snow 
load  should  be  used.  For  a  latitude  of  about  35  to  40  degrees, 
the  writer  recommends  a  minimum  snow  load  of  10  pounds, 
and  a  maximum  snow  load  of  20  pounds  per  square  foot  of  the 
horizontal  projection  of  the  roof;  the  former  to  be  used  in 
connection  with  a  maximum  wind  load,  and  the  latter  when 
the  wind  load  is  not  considered. 


Art.  3.  WIND    LOAD.  73 

ART.  3.    WIND  LOAD. 

92.  Wind  Pressure.     The  pressure  of  the  wind  against  a 
roof  surface  depends  upon  the  velocity  and  direction  of  the 
wind,   and   upon   the   inclination   of   the   roof.     The   wind   is 
assumed  to  move  horizontally,  and  the  pressure  against  a  flat 
surface,  normal  to  the  direction  of  the  wind,  may  be  found  by 
the  formula 

P  —  0.004  V2,  (4) 

where  P  is  the  pressure  in  pounds  per  square  foot  on  a  flat 
normal  surface,  and  V  is  the  velocity  of  the  wind  in  miles 
per  hour. 

The  pressure  on  a  flat  vertical  surface  is  usually  taken  at 
about  30  pounds  per  square  foot,  which  is  equivalent  to  a  wind 
velocity  of  87  miles  per  hour.  This  assumed  pressure  is  suffi- 
cient for  all  except  the  most  exposed  positions. 

93.  Wind   Pressure   on   Inclined    Surfaces.     The    normal 
wind  pressure  upon  an  inclined  surface  varies  with  the  inclina- 
tion of  the  roof;  and  several  formulae  have  been  derived  for 
finding  the  normal  component.     The  best  known  of  these  are 
Duchemin's,  Hutton's,  and  The  Straight  Line  formulae. 

Duchemin's  formula  is 

p  __  p_2_sin_A 
Fn~        i+sin2A' 

where     Pn  is  the  normal  component  of  the  wind  pressure, 
P  is  the  pressure  per  sq.  ft.  on  a  vertical  surface, 
A  is  the  angle  which  the  roof  surface  makes  with 

the  horizontal,  in  degrees. 
Hutton's  formula  is 

Pn  —  Psin  A,1-842'08*-1  (6) 

where  Pn,  P,  and  A  are  the  same  as  in  Duchemin's  formula. 
The  Straight  Line  formula  is 


45 
where  Pn,  P,  and  A  are  the  same  as  in  Duchemin's  formula. 


(7) 


LOADS. 


Chap.  VIII. 


Duchemin's  formula  is  based  upon  carefully  conducted 
experiments,  gives  larger  values,  and  is  considered  more  reli- 
able than  Hutton's.  The  Straight  Line  formula  is  preferred 
by  many  on  account  of  its  simplicity,  and  gives  results  which 
agree  quite  closely  with  experiments. 


0       5        10       15      20      25      50     .35       40      45       50     55     60" 
Angle, A,  Roof  Makes  with  Horizontal  in  Degrees. 

FIG.  47.     NORMAL  WIND  LOAD  ON  ROOF  BY  DIFFERENT  FORMULAE. 

Fig.  47  gives  values  for  the  normal  wind  pressure  Pn  in 
pounds  per  square  foot,  in  terms  of  the  pressure  on  a  vertical 
surface  and  of  the  angle  which  the  roof  surface  makes  with 
the  horizontal.  The  use  of  this  diagram  will  greatly  lessen 
the  work  required  to  find  the  normal  wind  pressure  when  the 
pressure  on  a  vertical  surface  and  the  inclination  of  the  roof 
are  known, 


CHAPTER  IX. 

KEACTIONS. 

The  reactions  are  the  forces  which  if  applied  at  the  center 
of  the  bearings  of  a  truss  would  hold  in  equilibrium  the  weight 
of  the  truss  and  the  loads  supported  by  it.  The  reactions  are 
numerically  equal  to  the  pressures  exerted  by  the  truss  against 
its  supports.  As  the  method  of  finding  the  reactions  for  vertical 
loads  differs  somewhat  from  that  for  inclined  loads,  the  determi- 
nation of  these  reactions  will  be  considered  in  separate  articles, 
the  first  article  treating  of  the  reactions  for  dead  and  snow  loads, 
and  the  second  of  the  reactions  for  wind  loads. 


ART.  i.     REACTIONS  FOR  DEAD  AND  SNOW  LOADS. 

Before  the  reactions  can  be  determined,  it  is  necessary  to 
find  the  loads  that  are  supported  by  the  truss.  The  purlins  are 
usually  placed  at  the  panel  points  of  the  upper  chord,  and  the 
loads  are  considered  to  act  at  these  panel  points.  The  method 
of  finding  the  joint  loads  and  the  dead  load  reactions  will  now 
be  shown  by  the  solution  of  the  following  problem. 

94.  Problem.  It  is  required  to  find  the  dead  load  reac- 
tions for  the  truss  shown  in  Fig.  48.  The  truss  has  a  span  of 
48  feet,  a  rise  of  12  feet,  and  the  distance  apart,  center  to  center 
of  trusses,  is  14  feet.  The  roof  is  of  slate,  laid  on  sheathing, 
which  is  supported  by  wooden  rafters.  The  purlins  and  the 
truss  are  of  steel. 

The  solution  will  be  divided  into  two  parts ;  the  determina- 
tion of  (a)  the  joint  loads,  and  (b)  the  reactions. 

75 


76 


REACTIONS. 


Chap.  IX. 


FIG.  48. 


(a)  Joint  Loads.   Referring  to  the  table  given  in  §  90  (i), 

it  is  seen  that  the  weight 
of  the  slate  covering  may 
be  taken  at  10  pounds,  and 
the  weight  of  the  sheath- 
ing at  3  pounds  per  square 
foot  of  roof  surface.  From 
§  90  (2),  it  is  seen  that  the  weight  of  the  purlins  and  bracing 
may  be  taken  at  3  pounds,  and  the  weight  of  the  rafters  at  2.5 
pounds  per  square  foot  of  roof  surface.  This  gives  a  total 
weight  of  18.5  pounds  per  square  foot  of  roof  surface,  exclusive 
of  the  weight  of  the  truss  itself. 

The  length  of  one-half  of  the  upper  chord  —V  24*  +  I22  feet 
=  26.83  feet-  This  length  is  divided  into  three  equal  parts  by 
the  web  bracing,  making  the  length  of  each  panel  of  the  upper 
chord  equal  to  26.83  -f-  3  =  8.94  feet.  Since  the  joint  loads  are 
taken  at  panel  points,  and  the  trusses  are  spaced  14  feet  apart, 
the  joint  load  supported  by  the  truss,  exclusive  of  the  weight 
of  the  truss,  =  8.94  X  14  X  18.5  =  2315  pounds. 

The  weight  of  the  roof  truss  per  square  foot  of  horizontal 
projection  for  a  capacity  P  of  40  pounds  (which  is  about  right 
for  the  given  roof  truss)  is  given  by  formula  (3),  §  90,  and  is 
about  3  pounds.  This  is  equivalent  to  a  joint  load  of  8  X  14  X 
3  =  336  pounds.  The  total  joint  load  is  therefore  equal  to  2315 
+  336  =  2651  pounds,  or  say  2650  pounds.  Referring  to  Fig. 
48,  it  is  seen  that  the  loads  acting  at  the  joints  B,  C,  D,  E, 
and  F  are  full  panel  loads;  while  those  acting  at  A  and  G 
support  but  half  the  area,  and  are  half  panel  loads.  The  loads 
acting  at  B,  C,  D,  E,  and  F  are  each  equal  to  2650  pounds ; 
while  those  acting  at  A  and  G  are  each  equal  to  1325  pounds. 

(b)  Reactions.     The  problem  now  is  to  find  the  reactions  at 
the  ends  of  the  truss  loaded  as  shown  in  Fig.  49,  or  in  other 
words,  it  is  required  to  find  the  two  forces  acting  at  the  ends 
of  the  truss,  which  will  hold  in  equilibrium  the  given  loads. 

Construct  the  force  polygon  ABCDEFGH  for  the  given 
loads  on  the  truss.  In  this  case,  the  force  polygon  is  a  straight 
line,  and  is  called  the  load  line.  Assume  any  pole  O,  construct 


Art.  1. 


DEAD    LOAD   REACTIONS. 


77 


the  funicular  polygon,  and  draw  the  closing  string  of  the  poly- 
gon. Then  the  dividing  ray,  drawn  through  the  pole  O  paral- 
lel to  this  closing  string,  will  determine  the  magnitudes  of  the 
two  reactions,  HM  being  the  magnitude  of  the  right  reaction 
R2,  and  MA  that  of  the  left  reaction  R±  (§  43).  The  lines  of 


0*        5000* 


A-r 
B 


R, 


FIG.  49.     DEAD  LOAD  REACTIONS. 


action  of  these  reactions  will  be  parallel  to  the  load  line — 
which  is  vertical.  By  scaling  the  lines  HM  and  MA,  it  is  found 
that  these  reactions  are  each  equal  to  7950  pounds. 

The  reactions  might  have  been  found,  algebraically,  by  tak- 
ing moments  about  the  supports.  Thus,  to  find  the  value  of 
the  left  reaction,  take  moments  about  the  right  support.  Then 
the  equation  of  moments  is  +  R  X  48 —  1325  X  48  —  2650  X 
40  —  2650  X  32  —  2650  X  24  —  2650  X  16  —  2650  X  8  =  o. 
Solving  this  equation  gives  Ra  =F=  7950.  Since  the  two  reac- 
tions must  hold  the  given  loads  in  equilibrium,  the  right  reac- 
tion is  therefore  equal  to  the  total  load  minus  the  left  reaction 
=  15  900  —  7950  =  7950  pounds. 

If  the  loads  are  symmetrical  with  respect  to  the  center  line 
of  the  truss,  it  is  evident  that  the  reactions  are  equal,  and  that 
each  is  equal  to  15  900  -=-  2  =  7950  pounds. 

95.  Snow  Load  Reactions.  The  reactions  and  the  stresses 
for  snow  loads  are  usually  considered  separately  from  those 
due  to  dead  loads.  The  same  methods  may  be  used  for  find- 
ing the  snow  load  reactions  as  have  been  explained  for  finding 
the  dead  load  reactions,  and  need  no  further  explanation. 


78  REACTIONS.  Chap.  IX. 

t 

96.  Effective  Reactions.  It  is  seen  from  Fig.  49  that  the 
half  joint  loads  at  the  ends  are  transferred  directly  to  the 
supports  without  causing  any  stress  in  the  truss.  These  half 
loads  act  through  the  same  lines  as  the  total  reactions,  and 
subtract  from  these  reactions.  The  resulting  forces  are  called 
the  effective  reactions.  If  the  effective  reactions  are  used,  the 
half  loads  at  the  ends  of  the  truss  are  neglected  in  computing 
the  stresses  in  the  truss.  The  effective  reactions  for  the  truss 
loaded  as  shown  in  Fig.  49  are  each  equal  to  7950 — 1325  = 
6625  pounds. 


ART.  2.     REACTIONS  FOR  WIND  LOADS. 

The  magnitudes  and  lines  of  action  of  the  wind  load  reac- 
tions for  any  given  truss  depend  upon  the  condition  of  the 
ends  of  the  truss.  If  the  span  is  small,  or  if  the  truss  is  made 
of  wood,  the  ends  are  usually  fixed  to  the  supports  by  anchor 
bolts.  For  large  spans,  the  changes  of  temperature  and  the 
deflections  due  to  the  loads  cause  the  truss  to  expand  or  con- 
tract a  considerable  amount.  If  the  ends  are  fixed,  the  tem- 
perature changes  and  the  loads  may  produce  large  stresses  in 
the  truss.  The  usual  method  of  providing  for  the  changes  in 
length  is  to  place  one  end  of  the  truss  upon  a  planed  base 
plate  or  upon  rollers. 

97.  Wind  Load  Reactions.     The  wind  load  reactions  will 
be  determined  for  each  of  the  following  assumptions:    (i)  that 
both  ends  of  the  truss  are  fixed ;   and  (2)  that  one  end  of  the 
truss  is  supported  upon  rollers.     In  the  example  that  will  be 
given,  a  triangular  form  of  truss  will  be  used,  although  the 
methods  employed  are  applicable  to  any  type  of  simple  truss. 

98.  (i)     Truss  Fixed  at  Both  Supports.     The  problem  of 
finding  the  lines  of  action  of  the  wind  load  reactions  for  a  truss 
fixed  at  both  ends  is  indeterminate ;   but  assumptions  may  be 
made  which  will  give  approximately  correct  results.    The  ver- 
tical   components   of   the    reactions   ate    independent   of   any 
assumptions ;  but  the  horizontal  components  depend  upon  the 


Art. 


WIND  LOAD  REACTIONS. 


79 


condition  of  the  ends  of  the  truss.  If  the  roof  is  comparatively 
flat,  i.  e.,  if  the  resultant  of  the  normal  components  of  the  wind 
loads  is  nearly  vertical,  the  reactions  may  be  taken  parallel  to 
this  resultant;  as  each  reaction  will  then  be  approximately 
vertical.  The  above  assumption  gives  erroneous  results  when 
the  roof  makes  a  large  angle  with  the  horizontal.  The  assump- 
tion that  the  horizontal  components  of  the  reactions  are  equal 
more  nearly  approximates  actual  conditions ;  and  this  is  par- 
ticularly true  if  the  ends  of  the  truss  are  fastened  to  the  sup- 
ports in  the  same  manner  and  the  supports  are  equally  elastic. 
The  method  of  finding  the  reactions  of  a  truss  with  fixed  ends 
will  now  be  shown  by  a  problem,  assuming  (a)  that  the  reac- 
tions are  parallel,  and  (b)  that  the  horizontal  components  of 
the  reactions  are  equal. 

(a)  Reactions  Parallel.  It  is  required  to  find  the  wind  load 
reactions  for  the  truss  shown  in  Fig.  48.  This  truss  has  a  span 
of  48  feet  and  a  rise  of  12  feet,  the  distance  between  trusses 
being  14  feet.  The  wind  pressure  on  a  vertical  surface  will  be 
taken  at  30  pounds  per  square  foot,  and  Duchemin's  formula 
will  be  used  for  finding  its  normal  component.  Referring  to 
Fig.  47,  it  is  seen  that,  for  P  =  30  pounds  and  for  a  pitch  of 
one-fourth,  the  component  normal  to  the  roof  is  about  22 
pounds.  The  length  of  a  panel  of  the  upper  chord  is  8.94  feet 
(see  §  94,  a).  The  panel  load  is  therefore  equal  to  8.94  X  14  X 
22  =  2753,  or  say  2750  pounds.  The  half  panel  loads  at  the 
ends  of  the  truss  and  at  the  apex  are  each  equal  to  1375  pounds. 


\  /-\  1^'  -?n-  ~~        — --^--A        n*      ?nnn"  Af\r 


10'  20 


FIG.   50.     WIND  LOAD  REACTIONS — ENDS  FIXED. 


80  REACTIONS.  Chap.  IX. 

The  truss  with  its  loads  is  shown  in  Fig.  50.  To  determine 
the  reactions,  construct  the  force  polygon  or  load  line  ABCDE, 
assume  a  pole,  and  draw  the  funicular  polygon.  The  dividing 
ray  OF,  drawn  through  the  pole  O  parallel  to  the  closing 
string  of  the  funicular  polygon,  will  give  the  magnitudes  of 
the  two  reactions,  FA  being  the  magnitude  of  the  left  reaction, 
and  EF  that  of  the  right.  The  lines  of  action  of  these  reac- 
tions are  given  by  drawing,  through  the  supports,  lines  paral- 
lel to  the  load  line  AE.  By  scaling  the  lines  FA  and  EF,  it  is 
seen  that  the  left  reaction  is  equal  to  5670  pounds,  and  the  right 
reaction  to  2580  pounds.  The  values  given  for  the  reactions 
were  actually  taken  from  a  diagram  four  times  as  large  as  that 
shown  in  Fig.  50;  as  the  scale  used  here  is  too  small  to  give 
accurate  results. 

(b)  Horizontal  Components  of  Reactions  Equal.  The  reac- 
tions will  be  found  for  the  same  truss  and  loads  as  in  Fig.  50, 
assuming  that  the  horizontal  components  of  the  reactions  are 
equal.  First  find  the  reactions,  as  in  §  98  (a),  assuming  that 
their  lines  of  action  are  parallel.  The  vertical  components  of 
these  reactions  are  correct ;  since  they  are  independent  of  the 
ends  of  the  truss.  Now  draw  a  horizontal  line  through  the 
point  F  (Fig.  50),  and  make  mn  equal  to  the  horizontal  com- 
ponent of  all  the  loads  on  the  truss.  This  may  be  done  by 
drawing  vertical  lines  throijgh  A  and  E,  intersecting  the  hori- 
zontal line  at  the  points  m  and  n,  respectively.  Bisect  mn, 
making  mF'  =  F'n,  and  draw  the  lines  F'A  and  EF',  which 
represent  the  magnitudes  of  the  required  reactions.  The  lines 
of  action  of  these  reactions  are  given  by  drawing  lines  through 
the  left  and  also  through  the  right  points  of  support  of  the 
truss  parallel  respectively  to  F'A  and  EF'.  By  scaling  the 
lines  F'A  and  EF',  it  is  seen  that  the  left  reaction  is  equal  to 
5400  pounds,  and  the  right  reaction  to  2960  pounds. 

99.  (2)  One  End  of  Truss  Supported  on  Rollers.  If  one 
end  of  the  truss  is  supported  on  rollers,  then  the  roller  end  can 
take  no  horizontal  component  of  the  wind  loads  (neglecting 
the  friction  of  the  rollers)  ;  for  there  would  be  a  continued 
movement  of  the  rollers  if  a  horizontal  force  was  applied  at 


Art. 


WIND   LOAD   REACTIONS. 


81 


this  end  of  the  truss.  The  reaction  at  the  roller  end  is  there- 
fore vertical,  and  hence  all  the  horizontal  component  of  the 
wind  loads  must  be  taken  up  at  the  fixed  end  of  the  truss. 
Since  the  wind  may  come  from  either  of  two  opposite  direc- 
tions, the  rollers  may  be  under  the  leeward  side,  or  under  the 
windward  side  of  the  truss.  The  method  for  finding  the  reac- 
tions will  now  be  shown  by  a  problem,  considering  (a)  that 
the  rollers  are  under  the  leeward  end  of  the  truss,  and  (b)  that 
the  rollers  are  under  the  windward  end. 

(a)  Rollers  Under  Leezvard  End  of  Truss.  It  is  required 
to  find  the  wind  load  reactions  for  the  truss  shown  in  Fig.  48, 
the  leeward  end  being  on  rollers.  The  wind  loads  are  the 
same  as  those  shown  in  Fig.  50.  In  this  problem,  the  line  of 
action  of  the  reaction  at  the  roller  end  is  vertical ;  while  the 
line  of  action  of  the  reaction  at  the  fixed  end  is  unknown,  its 
point  of  application,  only,  being  known.  To  determine  the 
reactions  for  this  case,  apply  the  method  explained  in  §  44. 


FIG.  51.     WIND  LOAD  REACTIONS — ONE  END  ON  ROLLERS. 

Draw  the  load  line  ABCDE  (Fig.  51)  for  the  given  loads, 
assume  any  pole  O,  and  draw  the  funicular  polygon  (shown  by 
full  lines),  starting  the  polygon  at  the  only  known  point  on  the 
left  reaction — its  point  of  application  at  the  left  end  of  the 
truss.  From  the  pole  O,  draw  the  dividing  ray  OF  parallel 
to  the  closing  string  of  the  funicular  polygon  to  meet  a  vertical 
line  drawn  through  E  parallel  to  the  known  direction  of  the 
right  reaction.  Connect  the  points  A  and  F ;  then  EF  repre- 
sents the  magnitude  of  the  right  reaction,  and  FA  that  of  the 


82  REACTIONS.  Chap.  IX. 

left  reaction.  The  lines  of  action  of  these  reactions,  which  are 
represented  by  R!  and  R2,  are  given  by  drawing  lines  through 
the  ends  of  the  truss  parallel  respectively  to  EF  and  FA  (§  44). 
By  scaling  the  lines  EF  and  FA,  it  is  seen  that  the  right  reac- 
tion is  equal  to  2310  pounds,  and  the  left  reaction  to  6270 
pounds. 

The  reactions  might  also  have  been  found  by  a  slightly  dif- 
ferent method,  as  follows :  Find  the  reactions  for  the  truss 
fixed  at  both  ends,  assuming  that  their  lines  of  action  are  par- 
allel (see  §  98  (a),  and  Fig.  50).  EF  and  FA  represent  the 
magnitudes  of  these  reactions,  their  vertical  components  being 
independent  of  the  condition  of  the  ends  of  the  truss.  Now 
since  the  right  reaction  can  have  no  horizontal  component, 
draw  the  line  En  through  E  to  meet  the  horizontal  line  through 
F;  then  En  represents  the  magnitude  of  this  reaction.  Now 
all  the  horizontal  component  of  the  wind  loads  must  be  taken 
up  at  the  left  end  of  the  truss,  this  component  being  repre- 
sented by  the  line  mn.  The  vertical  component  of  the  left 
reaction  is  represented  by  the  line  mA;  therefore  the  result- 
ant of  these  two  components,  or  the  line  nA  (not  drawn), 
represents  the  magnitude  of  the  left  reaction. 

(b)  Rollers  Under  Windward  End  of  Truss.  If  the  rollers 
are  under  the  windward,  instead  of  the  leeward  end  of  the 
truss,  then  the  left  reaction  is  vertical  and  the  right  is  inclined, 
the  only  known  point  on  the  right  reaction  being  its  point  of 
application  at  the  right  support.  For  this  case,  the  funicular 
polygon  must  start  at  the  right  support,  the  remainder  of  the 
solution  being  similar  to  that  explained  in  §  99  (a).  The  con- 
struction for  this  solution  is  shown  by  the  dotted  lines  in  Fig. 
51,  EF'  and  F'A  representing  the  magnitudes  of  the  two  reac- 
tions, their  lines  of  action  R2'  and  R/  passing  through  the 
right  point  of  support  and  the  left  point,  respectively.  By 
scaling  the  lines  EF'  and  F'A,  it  is 'seen  that  the  right  reaction 
is  equal  to  4350  pounds,  and  the  left  reaction  to  5070  pounds. 

The  reactions  might  also  have  been  found  by  a  method 
similar  to  that  explained  in  the  last  paragraph  of  §  99  (a),  mA 
and  Em  (not  drawn)  representing  these  reactions  (see  Fig.  50). 


COMPARISON    OF    REACTIONS. 


83 


100.  Fig.  52  shows  to  scale  the  wind  loads  and  the  wind 
load  reactions  for  different  assumptions  as  to  the  condition  of 
the  ends  of  the  truss.  It  is  seen  from  the  figure  that  the  verti- 


FIG.  52. 


Both  ends  fixed,  reactions  parallel,  shown  by 

Both  ends  fixed ,  hor.  com'ps  equal .  shown  by 

Windward  end  of  truss  on  rollers,  shown  by 

Leeward  end  of  truss  on  rollers. shown  by 

WIND  LOAD  REACTIONS  FOR  DIFFERENT  ASSUMPTIONS  AS  TO  CONDITION 
OF  ENDS  OF  TRUSS. 


cal  components  of  the  reactions  are  independent  of,  and  that 
the  horizontal  components  are  dependent  upon,  the  assump- 
tions as  to  the  condition  of  the  ends  of  the  truss.  If  the  roof 
had  made  a  greater  angle  with  the  horizontal  than  that  shown 
in  the  figure,  then  the  differences  in  the  reactions,  due  to  the 
assumptions  as  to  the  condition  of  the  ends  of  the  truss,  would 
have  been  more  apparent. 

101.  In  determining  the  reactions  for  a  truss,  great  care 
should  be  exercised  in  drawing  the  truss,  and  in  locating  the 
panel  points  accurately.  The  truss  and  diagrams  should  be 
drawn  to  a  large  scale  to  insure  good  results.  Care  should  be 
taken  in  laying  out  the  load  line,  and  the  pole  should  be  chosen 
in  such  a  position  that  acute  intersections  are  avoided. 


CHAPTER  X. 

STRESSES   IN   EOOF   TKUSSES. 

The  determination  of  the  external  forces  acting  upon  a  truss 
has  been  taken  up  in  Chapter  VIII  and  Chapter  IX.  The  dead 
and  wind  loads  acting  upon  the  roof  are  transferred  to  the 
supports  through  the  members  of  the  truss,  and  this  chapter 
will  treat  of  the  determination  of  the  stresses  in  the  members 
due  to  these  external  forces. 


ART.  i.     DEFINITIONS  AND  GENERAL  METHODS  FOR 
DETERMINING  STRESSES. 

102.  Definitions.  The  external  loads  tend  to  distort  the 
truss,  i.e.,  to  shorten  some  of  the  members,  and  to  lengthen 
others ;  and  since  the  materials  used  in  the  construction  of  the 
truss  are  not  entirely  inelastic,  the  members  are  actually  dis- 
torted. The  deformation  in  any  member  caused  by  the  loads 
is  called  strain,  and  the  internal  force  which  is  developed  in 
the  member  and  which  tends  to  resist  the  deformation,  or 
strain,  is  called  stress.  There  are  three  kinds  of  stress  which 
may  be  developed  by  the  external  forces,  viz. :  (a)  tension,  (b) 
compression,  and  (c)  shear. 

(a)  Tension.     A  member  is  subjected  to  a  tensile  stress,  or 
is  in  tension,  if  there  is  a  tendency  for  the  particles  of  the  mem- 
ber to  be  pulled  apart  in  a  direction  normal  to  the  surface  of 
separation.     In  this  case,  the  external  forces  causing  the  ten- 
sion act  in  a  direction  from  the  center  toward  the  ends;    and 
therefore  the  internal  forces,  or  stresses,  act  from  the  ends 
toward  the  center. 

(b)  Compression.     A  member  is  subjected  to  a  compressive 

84 


Art.  1.  METHODS    FOR   DETERMINING    STRESSES.  85 

stress,  or  is  in  compression,  if  there  is  a  tendency  for  the  par- 
ticles of  the  member  to  move  toward  each  other.  If  the  mem- 
ber is  in  compression,  the  external  forces  causing  the  compres- 
sion act  in  a  direction  from  the  ends  toward  the  center;  and 
therefore  the  internal  forces,  or  stresses,  act  from  the  center 
toward  the  ends. 

(c)  Shear.  A  member  is  subjected  to  a  shearing  stress, 
or  is  in  shear,  if  there  is  a  tendency  for  the  particles  of  the 
member  to  slide  past  each  other. 

Since  the  loads  are  generally  applied  at  the  joints,  the  mem- 
bers are  usually  subjected  to  a  longitudinal  stress  only,  and 
are  either  in  tension  or  in  compression ;  although  a  shearing 
stress  may  be  developed  at  the  connections  of  the  members. 

103.  General  Methods  for  Determining  Stresses.  It  has 
been  shown  in  the  preceding  chapter  that  the  reactions  of  a 
truss  may  be  determined,  graphically,  by  means  of  the  force 
and  the  funicular  polygons.  They  may  also  be  determined, 
algebraically,  by  the  three  fundamental  equations  of  equilib- 
rium : 

2  horizontal  components  of  forces  =  o,  (i) 

2  vertical  components  of  forces  =  o,  (2) 

2  moment  of  forces  about  any  point  =  o.  (3) 

Having  found  the  reactions  (see  Chapter  IX),  the  stresses 
may  be  determined  either  by  equations  (i)  and  (2)  or  by 
equation  (3).  The  first  two  equations  involve  the  resolution 
of  forces,  and  they  may  be  solved  either  algebraically  or 
graphically.  The  third  equation  involves  the  moments  of 
forces,  and  it  may  also  be  solved  either  algebraically  or  graph- 
ically. There  are,  therefore,  four  methods  for  determining  the 
stresses  in  a  truss. 

Moment  of  Forces: 

Algebraic  Method,        (a) 
Graphic  Method.  (b) 

Resolution  of  Forces: 

Algebraic  Method,        (c) 
Graphic  Method.          (d) 
It  is  possible  to  solve  the  stresses  in  the  members  of  any 


86  STRESSES   IN   ROOF   TRUSSES.  Chap.  X. 

simple  truss  by  using  any  one  of  the  above  four  methods ;  but 
all  are  not  equally  well  suited  to  any  particular  case.  It  is 
usually  the  case  that  a  certain  one  of  the  four  methods  is  bet- 
ter suited  to  a  particular  problem  than  are  any  of  the  other 
three ;  and  in  solving  stresses,  the  problem  should  be  studied 
in  order  that  the  simplest  solution  may  be  found. 

104.  Notation.     The  notation  that  will  be  used  to  desig- 

nate the  members  of  a  truss  is 
known  as  Bow's  notation,  and 
is  shown  in  Fig.  53.  Referring 
to  this  figure,  it  is  seen  that  the 
FIG.  53.  NOTATION.  upper  chord  members,  begin- 

ning at  the  left  end  of  the  truss,  may  be  designated  by  X-i,  X-2, 
X-4,  X-4',  X-2',  and  X-i';  the  lower  chord  members  by  Y-i, 
Y-3>  Y-5,  Y-5',  ¥-3',  and  Y-i';  and  the  web  members  by  1-2, 
2-3,  3-4,  4-5,  5-5',  5'-4',  4'-$',  3^2',  and  2'-:'.  The  numerical 
value  of  the  stress  will  generally  be  written  directly  on  the 
member. 

A  tensile  stress  will  be  denoted  by  prefixing  a  plus  (  +  )  sign, 
and  a  compressive  stress  by  prefixing  a  minus  ( — )  sign  before 
the  number  representing  the  stress.  This  use  of  these  signs  is 
not  universal,  but  the  above  designation  was  adopted  as  it  is  more 
often  employed. 

105.  It  is  the  object  of  this  work  to  deal  with  graphic— 
rather  than  with  algebraic  methods ;   and  since  the  method  of 
graphic  moments  is  not  well  suited  to  the  determination  of 
the  stresses  in  a  truss,  except  to  explain  other  methods,  the 
first  three  methods  will  be  treated  only  briefly,  while  this  text 

,  will  deal  chiefly  with  the  determination  of  stresses  by  the 
fourth  method — graphic  resolution.  These  four  methods  will 
now  be  taken  up  in  the  order  shown  in  §  103. 

ART.  2.     STRESSES  BY  ALGEBRAIC  MOMENTS. 

The  method  of  algebraic  moments  furnishes  a  convenient 
means  of  finding  stresses,  especially  when  the  upper  and  lower 
chords  of  the  truss  are  not  parallel. 


Art. 


STRESSES    BY    ALGEBRAIC    MOMENTS. 


87 


1 06.  Method  of  Computing  Stresses  by  Algebraic  Mo- 
ments. To  obtain  the  stress  in  any  particular  member,  cut 
the  truss  by  a  section;  and  replace  the  stresses  in  the  mem- 
bers cut,  by  external  forces.  These  forces  are  equal  to  the 
stresses  in  the  members  cut,  but  act  in  an  opposite  direction. 
The  section  should  be  so  taken  that  the  member  whose  stress 
is  required  is  cut  by  it;  and  if  possible,  so  that  the  other 
members  cut  by  the  section  (excepting  the  one  whose  stress 
is  required)  pass  through  a  common  point,  which  point  is 
taken  as  the  center  of  moments.  To  determine  the  sign  of 
the  moment  and  of  the  resulting  stress,  assume  the  unknown 
external  force  to  act  away  from  the  cut  section,  i.  e.,  to  cause 
tension.  Write  the  equation  of  moments,  considering  the 
external  forces  which  replace  the  members  cut  and  those  on 
one  side  of  the  section  only,  equate  to  zero  (see  §  103,  equa- 
tion 3),  and  solve  for  the  unknown  stress.  The  sign  of  the 
result  will  then  indicate  the  kind  of  stress.  If  it  is  positive, 
the  assumed  direction  of  the  unknown  force  is  correct,  and 
the  stress  is  tension.  If  the  result  is  negative,  the  assumed 
direction  is  incorrect,  and  the  stress  is  compression,  i.  e.,  the 


(b)  (c)  (d) 

FIG.  54.     STRESSES  BY  ALGEBRAIC   MOMENTS. 


88  STRESSES  IN  ROOF  TRUSSES.  ChaP-  X. 

equation  of  moments  is  not  equal  to  zero  unless  the  assumed 
direction  of  the  external  force  is  reversed. 

The  application  of  the  method  of  algebraic  moments  to  the 
determination  of  the  stresses  in  a  truss  will  now  be  shown. 

107.  Problem.  It  is  required  to  find  the  stresses,  due  to 
the  dead  load,  in  the  members  of  the  truss  shown  in  Fig.  54. 

To  determine  the  stress  in  the  member  X-i,  cut  the  mem- 
bers X-i  and  Y-i  by  the  section  m-m  (see  Fig.  54,  a),  replace 
these  members  by  external  forces,  assume  that  the  unknown 
external  force  replacing  the  required  stress  X-i  acts  away 
from  the  cut  section,  and  take  moments  about  the  joint  Lx 
(Fig.  54,  a).  Then 

+  R,  X  d  +  X-i  X  a  =  o, 

or  X-i  =  —  _  .  _  (compression).  (i) 

a 

Since  the  sign  of  the  result  is  negative,  the  equation  shows 
that  the  external  force  acts  in  an  opposite  direction  to  that 
assumed,  i.  e.,  that  it  acts  in  the  direction  shown  in  Fig.  54,  b, 
and  causes  compression. 

To  find  the  stress  in  the  member  Y-i,  use  the  same  section 
and  take  moments  about  Ulf  assuming  that  the  external  force 
replacing  the  stress  in  Y-i  acts  away  from  the  cut  section. 
Then 

+  RX  X  d  —  Y-i  X  e=o, 

or  Y-i  =  +  Rl  X  d  (tension).  (2) 

e 

Since  the  result  is  positive,  the  equation  shows  that  the 
external  force  acts  in  the  direction  assumed,  i.  e.,  that  it  acts 
in  the  direction  shown  in  Fig.  54,  b. 

To  find  the  stress  in  the  member  1-2,  cut  the  members 
X-2,  1-2,  and  Y-i  by  the  section  s-s  (see  Fig.  54,  c),  and  take 
moments  about  L0,  assuming  that  the  external  force  which 
replaces  the  stress  in  1-2  acts  away  from  the  cut  section.  Then 


or        1-2  =  —  _    =  —  P    (compression).    (3) 


Art.  2.  STRESSES   BY   ALGEBRAIC    MOMENTS.  89 

To  find  the  stress  in  X-2,  cut  the  members  X-2,  2-3,  and 
¥-3  by  the  section  n-n  (see  Fig.  54,  a  and  Fig.  54,  d),  and 
take  the  moments  about  Lx,  the  intersection  of  2-3  and  ¥-3. 
Then 

+  R1Xd  +  X-2Xa=o, 

RI  X  d 

or  X-2  =  — — * -  (compression).  (4) 

a 

To  determine  the  stress  in  2-3,  cut  the  members  by  the 
section  n-n,  assume  the  external  force  replacing  the  stress  in 
2-3  to  act  away  from  the  cut  section,  and  take  the  moments 


about  L0.   Then 


_|_pXd  — 2-3X0  =  0, 


P  X  d 

or  2-3  =  +  . .  (tension).  (5) 

c 

This  equation  shows  that  the  assumed  direction  of  the 
external  force  replacing  2-3  was  correct,  i.  e.,  that  this  force 
acts  away  from  the  cut  section,  as  shown  in  Fig.  54,  d. 

For  the  stress  in  ¥-3,  cut  the  members  by  the  section  n-n, 
and  take  moments  about  U2.  Then 

+  R±  X  2d  —  P  X  d  —  Y-3  X  f  =  o, 

+  R±  X  2d  —  P  X  d  (f.. 

or  Y-3=J _ —  (tension).  (6) 

For  the  stress  in  3-4,  take  a  section  (not  shown  in  the 
figure)  cutting  X~4,  3-4,  and  Y-3,  and  take  moments  about  L0. 
Then 

+  P  X  d  +  P  X  2d  +  3-4  X  2d  =o, 

—  P  X  d  —  P  X  2d  - 

or         3-4=  — j —        -=___J_  P    (compression).    (7) 

For  the  stress  in  X-4,  take  the  section  p-p,  and  the  center 
of  moments  at  L2.  Then 

+  RA  X  2d  —  P  X  d  +  X-4  X  b  =  o, 

-RtX2d  +  PXd 
or  X-4  = — —  -   (compression).     (8) 

For  the  stress  in  4-5,  take  the  section  p-p,  and  the  center 
of  moment  at  L0.  Then 


90  STRESSES  IN  ROOF  TRUSSES.  Chap.  X. 


+      X     +      X2        +  3  Pd 
or  4-5  =  —       -     —  =—  (tension).          (9) 

o  & 

For  the  stress  in  Y~5,  take  the  section  p-p,  and  the  center 
of  moments  at  U3.  Then 

+  R1X3d  —  PX2d  —  PXd  —  Y-5Xh  =  o,    orY-5  = 


h  h 

The  stress  in  the  member  5-5'  =  o,  as  may  be  shown  by 
taking  a  circular  section  cutting  Y~5,  5-5',  and  ¥-5',  and  the 
center  of  moments  at  L2. 

Since  the  truss  and  the  loads  are  symmetrical  about  the 
center  line,  it  is  evident  that  it  is  necessary  to  find  the  stresses 
for  only  one-half  of  the  truss. 

In  some  trusses,  it  is  impossible  to  take  a  section  so  that 
all  of  the  members  except  the  one  whose  stress  is  required  will 
pass  through  the  center  of  the  moments.  If  another  mem- 
ber cut  by  the  section  does  not  pass  through  the  center  of  the 
moments,  it  is  necessary  to  first  solve  for  the  stress  in  this 
member,  and  then  replace  this  stress  by  an  external  force.  If 
the  stress  in  this  member  is  tension,  the  external  force  is 
taken  as  acting  away  from  the  cut  section  ;  and  if  it  is  com- 
pression, toward  the  section. 

If  the  moment  arms  for  the  forces  are  computed  alge- 
braically, considerable  work  is  required  ;  and  these  arms  may 
generally  be  most  easily  found  by  drawing  the  truss  to  a 
large  scale,  and  scaling  the  moment  arms  from  the  diagram. 

One  of  the  most  important  advantages  of  the  method  of 
moments  is  that  the  stress  in  any  particular  member  may  be 
found  independently  of  that  in  any  other  member. 

108.  By  noting  the  results  obtained  for  the  stresses  in  the 
preceding  problem,  several  conclusions  may  be  drawn  as  to 
the  nature  of  the  stresses  in  the  different  members  of  the  truss. 
Referring  to  equations  (i)  and  (4),  §  107,  it  is  seen  that  the 
stress  in  X-i  is  equal  to  that  in  X-2  ;  and  referring  to  equa- 
tion (3),  it  is  seen  that  1-2  is  an  auxiliary  member  whose  func- 


4rt.  3.  STRESSES    BY    GRAPHIC    MOMENTS.  91 

tion  is  to  transfer  the  load  P  to  the  joint  Lx  (by  compression). 
Further,  if  there  is  no  load  at  L\,  the  stress  in  1-2  is-zero. 
Since  the  load  is  transferred  to  Lx  by  the  member  1-2,  the 
stress  in  X-i  must  be  equal  to  that  in  X-2;  as  the  load  has 
no  horizontal  component  and  all  of  its  vertical  component  is 
taken  up  by  the  member  1-2.  There  can  be  no  stress  in  the 
member  5-5'  unless  there  is  a  load  at  L3;  and  if  there  is  a 
load  at  that  point,  the  stress  in  5-5'  is  tension,  and  is  equal  to 
the  load.  The  member  5-5'  is  usually  put  in,  its  functions 
being  merely  to  support  the  lower  chord  and  prevent  deflec- 
tion. 

ART.  3.     STRESSES  BY  GRAPHIC  MOMENTS. 

The  stresses  in  the  members  of  a  truss  may  be  found  by 
graphic  moments,  although  this  method  is  not  generally  the 
simplest  that  may  be  used. 

109.  Stresses  by  Graphic  Moments.  This  method  is  some- 
what similar  to  that  of  algebraic  moments,  as  explained  in 
Art.  2 ;  except  that  in  this  method,  the  moment  of  the  exter- 
nal forces  is  found  graphically  instead  of  algebraically.  Since 
the  structure  is  in  equilibrium,  the  moment  of  the  known 
external  forces  must  be  equal  to  the  moment  of  the  forces 
which  replace  the  stresses  in  the  members  cut  by  the  section. 
If  all  the  members  cut,  except  one,  pass  through  the  center  of 
moments;  then  the  moment  of  the  external  force  replacing 
the  stress  in  this  member  must  be  equal  to  the  moment  of 
the  known  external  forces.  To  determine  the  stress  in  any 
particular  member,  cut  the  member  by  a  section,  and  take 
moments  about  the  intersection  of  the  other  members  cut. 
To  find  the  moment  of  the  known  external  forces,  construct 
the  force  and  the  funicular  polygons  for  these  forces.  Then 
the  moment  is  equal  to  the  pole  distance  multiplied  by  the  inter- 
cept. This  intercept  is  measured  on  a  line  drawn  through  the 
center  of  moments  parallel  to  the  resultant  of  the  external 
forces  on  one  side  of  the  section ;  and  is  the  distance  on  this 
line  cut  off  by  the  strings  drawn  parallel  to  the  rays  meeting 


92 


STRESSES  IN  ROOF  TRUSSES. 


Chap.  X. 


on  the  extremities  of  the  line  representing  the  magnitude  of 
the  resultant  in  the  force  polygon.  If  all  the  members  cut  by 
the  section,  except  the  one  whose  stress  is  required,  pass 
through  the  center  of  moments,  the  algebraic  sum  of  the 
moments  of  the  unknown  force  replacing  the  stress  in  this 
member  and  of  the  known  external  forces  on  one  side  of  the 
section  must  be  equal  to  zero. 

The  solution  of  the  following  problem  shows  the  applica- 
tion of  the  above  method  to  the  determination  of  the  stresses 
in  the  members  of  a  truss. 

no.  Problem.  It  is  required  to  find  the  stresses  in  the 
members  of  the  truss  loaded  as  shown  in  Fig.  55. 


P       X 


izf-: 


L_i 

n   /^    \ 

L-3 

V>9N 

/  :      P    ; 

V 

-d-i— 

m 

^ 

_/d  ^_d_^ 

N 
X 

r 
y^ 

If       : 

! 

— 

s 
,' 

**  "X 

^ 

I  y*        | 

Vs 

t 
^ 

,' 

Ra 


L-I-^-^l0 


-7^-H 


FIG.  55.     STRESSES  BY  GRAPHIC  MOMENTS. 

To  determine  the  stress  in  the  member  X-i,  cut  the  mem- 
ber by  the  section  m-m,  and  take  the  center  of  moments  at  Lt. 
Then  the  moment  of  the  left  reaction  is  equal  to  +  H  X  v3, 


Art.  3.  STRESSES    BY    GRAPHIC    MOMENTS.  93 

and  the  equation  of  moments  (assuming  the  unknown  external 
force  to  act  away  from  the  cut  section)  is 


A  plus  sign  placed  before  the  stress  indicates  tension,  and 
a  minus  sign  indicates  compression. 

To  find  the  stress  in  Y-i,  cut  the  member  by  the  section 
m-m,  and  take  the  center  of  moments  at  Uj.  Then 


To  find  the  stress  in  1-2,  take  a  section  (not  shown  in  the 
figure)  cutting  X-2,  1-2,  and  Y-i,  and  take  the  center  of 
moments  at  L0.  Then 

+  HXy1  +  i-2Xd=o, 

HXy, 
or  1-2  *.  --  -j—  .  (3) 

To  find  the  stress  in  X-2,  take  the  section  n-n,  and  the 
center  of  moments  at  Lx.  Then 

+  HXy8  +  X-2Xa=o, 

HXyg 
or  X-2  =  --  -  —  .  (4) 

cl 

To  find  the  stress  in  2-3,  take  the  section  n-n,  and  the' 
center  of  moments  at  L0.  Then 

+  HXyi  —  2-3  Xc  =o, 

HXyi 
or  2-3  —  +  -  -  —  .  (5) 

To  find  the  stress  in  Y~3,  take  the  section  n-n,  and  the 
center  of  moments  at  U2.  Then 


Y-3  =  +-.  (6) 


94  STRESSES  IN  ROOF  TRUSSES.  Chap.  X. 

To  find  the  stress  in  3-4,  take  a  section  (not  shown  in  the 
figure),  and  the  center  of  moments  at  L0.   Then 
+  HXya  +  3-4X2d  =  o, 

HX  YO 
3-4--— J-1.  (7) 

To  find  the  stress  in  X-4,  take  the  section  p-p,  and  the 
center  of  moments  at  L2.  Then 

+  HXy4  +  X-4Xb  =  o, 

or  X-4  = jpX  (8) 

To  find  the  stress  in  4-5,  take  section  p-p,  and  the  center 
of  moments  at  L0.  Then 

+  HXy2  —  4-5Xg=o, 
HXy3 

or  4-5  =  +  — .  (9) 

•  o 

To  find  the  stress  in  Y~5,  take  the  section  p-p,  and  the 
center  of  moments  at  U3.  Then 

+  HXy5  —  Y-5Xh=o, 
or  Y-5  =  +      ~-.  (10) 

Referring  to  the  equations  of  moments  given  in  §  no,  it 
is  seen  that  the  sign  of  the  moment  of  the  external  forces  to 
the  left  of  any  section  is  always  positive.  Keeping  this  in 
mind,  it  is  possible  to  write  the  value  for  the  stress  in  any 
member,  together  with  its  proper  sign,  without  first  writing 
the  equation  of  moments.  The  stress  in  any  particular  mem- 
ber— irrespective  of  whether  it  is  tension  or  compression — is 
equal  to  the  moment  of  the  external  forces  to  the  left  of  the 
section  cutting  the  member  divided  by  the  arm  of  the  force 
which  replaces  the  unknown  stress  in  the  member.  The  sign 
of  this  stress  is  opposite  to  that  of  the  moment  of  the  unknown 
external  force  replacing  the  stress  in  the  member,  assuming 
this  stress  to  act  away  from  the  cut  section,  i.  e.,  if  the  sign 
of  the  moment  of  this  external  force  is  negative,  the  stress  is 


Art.  4. 


STRESSES   BY    ALGEBRAIC    RESOLUTION. 


95 


tension ;   and  if  the  sign  is  positive,  the  stress  is  compression. 
It  is  seen  that  this  follows  directly  from  equations  of  moments. 


ART.  4.     STRESSES  BY  ALGEBRAIC  RESOLUTION. 

in.  Stresses  by  Algebraic  Resolution.  The  stresses  in 
the  members  of  a  truss  may  be  found  by  the  application  of 
equations  (i)  and  (2),  §  103,  which  are 

3    horizontal  components  of  forces  =  o,         ( i ) 
2<   vertical  components  of  forces     =o.         (2) 

The  above  equations  may  be  applied  either  (a)  to  the 
forces  at  a  joint,  or  (b)  to  the  forces  on  one  side  of  a  section, 
including  those  replacing  the  stresses  in  the  members  cut  by 
the  section.  The  method  of  algebraic  resolution  is  not  applica- 
ble if  more  than  two  of  the  forces  at  a  joint  or  at  a  section  are 
unknown ;  since  there  are  but  two  fundamental  equations  of 
resolution.  As  it  is  necessary  to  find  the  stresses  in  some  of 
the  members  before  those  in  other  members  may  be  found, 
this  method  may  be  most  easily  explained  by  solving  a  par- 
ticular problem  rather  than  a  general  one. 

In  writing  the  equations,  forces  acting  upward  and  to  the 
right  will  be  considered  positive,  and  those  acting  downward 
and  to  the  left,  negative. 


Sin  60°=  0.866  Cos.60°=0.500 

FIG.  56. 

v    (a)     Forces  at  a  Joint.     To  find  the  stresses  in  the  members 
meeting  at  a  joint,  apply  equations  (i)  and  (2),  §  in,  assum- 


96  STRESSES   IN    ROOF   TRUSSES.  Chap.  X. 

ing  that  the  unknown  forces  replacing  the  stresses  act  away 
from  the  joint,  i.  e.,  that  they  cause  tension.  If  all  the  forces 
are  known  except  two,  these  may  be  found  by  solving  the 
above  equations.  The  signs  of  the  results  will  determine  the 
kind  of  stress,  i.  e.,  a  plus  sign  indicates  that  the  assumed 
direction  is  correct  and  that  the  stress  is  tension,  while  a 
minus  sign  indicates  that  the  assumed  direction  is  incorrect 
and  that  the  stress  is  compression. 

Problem.  It  is  required  to  find  the  stresses  in  the  members 
of  the  truss  loaded  as  shown  in  Fig.  56,  all  loads  being  given 
in  pounds. 

To  find  the  stresses  in  X-i  and  Y-i,  apply  equations  (i) 
and  (2)  to  the  forces  at  the  joint  L0,  assuming  that  these 
forces  act  away  from  the  joint.  Then 

+  X-i  sin  60°  +  Y-i  =o, 
and  +  X-i  cos  6°°  +  3000  =  o. 

Substituting  the  values  of  sin  60°  (0.866)  and  cos  60° 
(0.500),  and  solving  the  equations  for  X-i  and  Y-i,  we  have 
X-i  =  —  6000  (compression),  and  Y-i  =  +  5196  (tension). 

To  find  the  stresses  in  X-2  and  1-2,  apply  equations  (i) 
and  (2)  to  the  forces  at  the  joint  U^  Then 

+  X-i  sin  60°  +  X-2  sin  60°  +  1-2  cos  60°  =  o, 
and     +  X-i  cos  60°  +  X-2  cos  60°  —  1-2  sin  60°  —  2000  =  o. 

Substituting  the  values  of  sin  60°  and  cos  60°,  also  the 
value  already  found  for  X-i,  and  solving  these  equations  for 
1-2  and  X-2,  we  have 

1-2  =  — 1732,  and  X-2  =  —  5000. 

To  find  the  stresses  in  2-3  and  Y~3,  apply  equations  (i) 
and  (2)  to  the  forces  at  the  joint  L^  Then 

—  Y-i  +  1-2  cos  60°  +  2-3  cos  60°  +  Y-3  =  o, 
and  — 1-2  sin  60°  +  2-3  sin  60°  =  o. 

Substituting  the  values  of  sin  60°  and  cos  60°,  also  the 
values  of  Y-i  and  1-2,  and  solving  these  equations  for  2-3 
and  Y-3,  we  have 

2-3  =  +  1732,  and  Y-3  =  +  3464. 


Art.  4. 


STRESSES   BY   ALGEBRAIC    RESOLUTION. 


97 


Fig.  56,  a,  shows  the  truss  just  solved,  together  with  its 
loads  and  stresses.  This  figure  illustrates  the  method  of  writ- 
ing the  stresses  on  the  members  of  the  truss. 


+  5196 

Sin  60°=0.866 


+  5464 
Y 

FIG.  56,  a. 


+5196 
Cos  60°=  0.500 


(b)  Forces  on  One  Side  of  a  Section.  The  method  of 
algebraic  resolution  may  also  be  applied  to  the  forces  on  one 
side  of  a  section,  including  those  replacing  the  stresses  in  the 
members  cut  by  the  section.  Since  the  forces  replacing  the 
stresses  in  the  members  cut  and  the  external  forces  on  one 
side  of  a  section  must  be  in  equilibrium,  we  may  apply  the 
two  fundamental  equations  (i)  and  (2),  §  103,  and  find  the 
unknown  stresses,  provided  not  more  than  two  of  the  stresses 
are  unknown.  The  kind  of  stress  may  be  determined  by 
assuming  the  unknown  external  forces  replacing  the  stresses 
in  the  members  cut  to  act  away  from  the  section.  A  plus 


FIG.  57. 


98  STRESSES  IN  ROOF  TRUSSES.  Chap.  X. 

sign   for  the  result  indicates  tension,   and   a  minus  sign,   com- 
pression. 

Problem.  It  is  required  to  find  the  stresses  in  the  mem- 
bers of  the  truss  loaded  as  shown  in  Fig.  57,  all  loads  being 
given  in  pounds. 

To  find  the  stresses  in  the  members  X-i  and  Y-i,  take  the 
section  m-m  (Fig.  57),  and  assume  that  the  unknown  forces 
replacing  the  stresses  in  X-i  and  Y-i  act  away  from  the  cut 
section.  If  a  known  stress  is  considered  in  any  equation,  the 
external  force  replacing  it  is  taken  as  acting  in  its  true  direc- 
tion. Then 

+  X-i  sin  60°  +  Y-i  =o, 
and  +  X-i  cos  60°  +  6000  =  o. 

Substituting  the  values  of  sin  60°  (0.866)  and  cos  60° 
(0.500)  in  these  equations,  and  solving  for  X-i  and  Y-i,  we 
have 

X-i  =  —  12  ooo,        and        Y-i  =  +  Jo  392. 

To  find  the  stresses  in  1-2  and  X-2,  take  the  section  n-n, 
and  assume  that  the  unknown  external  forces  replacing  the 
stresses  in  X-2  and  1-2  act  away  from  the  section  (the  direc- 
tion of  the  force  replacing  the  stress  in  Y-i  has  already  been 
found  to  act  away  from  the  section).  Then 

+  X-2  sin  60°  +  1-2  cos  60°  +  Y-i  =  o, 
and  +  X-2  cos  60°  —  1-2  sin  60°  +  6000  —  4000  =  o. 

Substituting  the  values  of  sin  60°  and  cos  60°,  also  the 
value  already  found  for  Y-i,  and  solving  these  equations  for 
1-2  and  X-2,  we  have 

1-2  =  —  3464,  and  X-2  =  —  10  ooo. 

To  find  the  stresses  in  2-3  and  ¥-3,  take  the  section  p-p, 
and  assume  that  the  unknown  external  forces  replacing  the 
stresses  in  2-3  and  ¥-3  act  away  from  the  cut  section  (the 
force  replacing  the  stress  in  X-2  has  already  been  found  to  act 
towards  the  section).  Then 

-  X-2  sin  60°  +  2-3  cos  60°  +  ¥-3  =  o, 
and  —  X-2  cos  60°  +  2-3  sin  60°  +  6000  —  4000  =  o 


Art.  5.  STRESSES  BY  GRAPHIC  RESOLUTION.  99 

Substituting  the  values  for  sin  60°  and  cos  60°,  also  the 
value  already  found  for  X-2,  and  solving  for  2-3  and  ¥-3,  we 
have 

'     2-3  =  +  3464,  and  ¥-3  =  +  6928. 


ART.  5.     STRESSES  BY  GRAPHIC  RESOLUTION. 

The  method  of  graphic  resolution  is  usually  the  most  con- 
venient one  for  finding  the  stresses  in  roof  trusses ;  since  it  is 
rapid  and  has  the  advantage  of  furnishing  a  check  on  the 
work.  It  consists  of  the  application  of  the  principle  of  the 
force  polygon  to  the  external  forces  and  stresses  acting  at 
each  joint  of  the  truss.  Since  the  external  forces  and  stresses 
at  each  joint  are  in  equilibrium,  the  force  polygon  must  close, 
and  the  forces  must  act  in  the  same  direction  around  the 
polygon  (see  §  28).  As  the  lines  of  action  of  all  the  forces 
are  known,  it  follows  that  if  a  sufficient  number  of  the  forces 
at  a  joint  are  completely  known  to  permit  of  the  drawing  of  a 
closed  polygon,  the  magnitudes  and  directions  of  the  unknown 
stresses  may  be  determined. 

The  reactions  may  be  found  by  means  of  the  force  and 
funicular  polygons,  as  explained  in  Chap.  II.  As  soon  as 
these  reactions  are  determined,  all  the  external  forces  acting 
on  the  truss  are  known,  and  the  stresses  may  then  be  found. 

112.  Stresses  by  Graphic  Resolution.  Loads  on  Upper 
Chord.  It  is  required  to  find  the  stresses  in  the  truss  loaded 
as  shown  in  Fig.  58. 

The  reactions  Rj  and  R2  are  first  found  by  constructing  the 
force  polygon  (Fig.  58,  b)  for  the  given  loads  (see  §  94,  b). 
If  both  the  truss  and  the  loads  are  symmetrical  about  the 
center  line,  it  is  unnecessary  to  construct  the  funicular  polygon ; 
as  each  reaction  is  equal  to  one-half  of  the  total  load.  The 
stresses  may  then  be  determined  by  applying  the  principle 
of  the  force  polygon  to  each  joint  of  the  truss.  Referring  to 
the  joint  L0  (Fig.  58,  a),  of  the  three  forces  acting  at  this 
joint,  it  is  seen  that  the  reaction  Rt  is  completely  known, 


100 


STRESSES  IN  ROOF  TRUSSES. 


Chap.  X. 


while  the  internal  forces  or  stresses  X-i  and  Y-i  are  unknown 
in  magnitude  and  direction.  These  forces  are  shown  in  Fig. 
58,  c.  The  force  polygon  for  these  forces  is  constructed  by 


K'ff 


41 
(f)  Joint  U2 


A- 
(h)  Joint  U3 


( i )  Stress  Diaqram 


FIG.  53.     STRESSES  BY  GRAPHIC  RESOLUTION. 


Art-  5-  STRESSES  BY  GRAPHIC  RESOLUTION.  101 

drawing  Rx  (Fig.  58,  c),  acting  upward,  equal  and  parallel  to 
the  known  left  reaction;  and,  from  the  extremities  X  and  Y 
of  this  line,  drawing  the  lines  X-i  and  Y-i  parallel  respect- 
ively to  the  members  of  the  truss  X-i  and  Y-i,  meeting  at 
the  point  I.  Then  X-i  and  Y-i  represent  to  scale  the  magni- 
tudes of  the  stresses  in  the  members  X-i  and  Y-i.  These 
forces  are  in  equilibrium,  so  they  must  act  around  the  polygon 
in  the  same  direction ;  and  by  applying  the  forces  to  the  joint 
L0,  it  is  seen  that  the  stress  X-i  acts  towards  the  joint  and  is 
compression,  while  the  stress  Y-i  acts  away  from  the  joint 
and  is  tension. 

The  forces  at  the  joint  U,.  are  next  taken,  instead  of  those 
at  Lj ;  since  at  the  former  there  are  but  two  unknown  forces, 
while  at  the  latter  there  are  three.  The  forces  at  U1  are  shown 
in  Fig.  58,  d.  Since  the  stress  X-i  has  been  found  to  be  com- 
pression, it  must  act  towards  the  joint  U^  The  force  P  is  also 
known,  while  the  internal  forces  X-2  and  1-2  are  known  in 
lines  of  action  only.  The  force  polygon  for  these  forces  (Fig. 
58,  d)  is  constructed  by  drawing  X-i,  acting  upward  and  to 
the  right,  equal  and  parallel  to  the  known  force  X-i,  and  the 
force  P,  acting  downward,  equal  and  parallel  to  the  known 
load  P.  The  force  polygon  is  then  closed  by  drawing  from 
the  point  X  the  line  X-2  parallel  to  the  member  X-2,  and 
from  the  point  I,  by  drawing  1-2  parallel  to  the  member  1-2. 
Then  X-2  and  1-2  represent  the  magnitudes  of  the  unknown 
forces,  X-2  acting  downward  and  to  the  left,  and  1-2  acting 
upward.  Both  X-2  and  1-2  are  compression;  as  may  be  seen 
by  applying  these  forces  to  the  joint  U1  (Fig.  58,  a). 

The  forces  acting  at  the  joint  Lj  are  next  considered. 
These  forces  are  shown  in  Fig.  58,  e,  the  forces  2-3  and  Y~3 
being  unknown.  The  unknown  forces  are  found  by  construct- 
ing the  force  polygon,  as  shown  in  Fig.  58,  e.  Applying  the 
forces  in  this  polygon  to  the  joint  Lu  it  is  seen  that  both  2-3 
and  Y— 3  act  away  from  the  joint,  and  therefore  the  stresses 
are  tension. 

The  forces  acting  at  the  joint  U2  are  shown  in  Fig.  58,  f. 
Of  these,  the  forces  P,  X-2,  and  2-3  are  known,  while  3-4  and 


102  STRESSES  IN  ROOF  TRUSSES.  Chap.  X. 

X-4  are  unknown.  The  unknown  forces  are  found  by  con- 
structing the  force  polygon  shown  in  Fig.  58,  f.  Applying  the 
unknown  forces  to  the  joint  U2,  it  is  seen  that  both  3-4  and 
X~4  are  compression. 

The  forces  acting  at  L2  are  shown  in  Fig.  58,  g.  The 
unknown  forces  4-5  and  Y~5  are  found  by  constructing  the 
force  polygon  shown  in  Fig.  58,  g.  Both  forces  act  away  from 
the  joint,  and  therefore  the  stresses  are  tension. 

The  forces  acting  at  U3,  together  with  their  force  polygon, 
are  shown  in  Fig.  58,  h.  The  stress  in  X-4'  is  compression, 
while  that  in  4/>~5/  is  tension.  The  stress  in  5-5'  is  zero. 

Since  the  truss  and  loads  are  symmetrical  about  the  center 
line,  the  force  polygons  for  the  joints  to  the  right  of  the  center 
need  not  be  drawn ;  as  they  are  the  same  as  those  for  the  left. 

Stress  Diagram.  Referring  to  the  separate  force  polygons 
shown  in  Fig.  58,  it  is  seen  that  some  of  the  forces  in  one 
polygon  are  repeated  in  another,  i.  e.,  it  is  necessary  to  find 
some  of  the  stresses  in  one  polygon  and  use  these  stresses  in 
drawing  the  next  polygon.  It  is  thus  seen  that  these  separate 
force  polygons  may  be  grouped  together  in  such  a  manner 
that  none  of  the  lines  are  drawn  twice.  Again  referring  to  the 
separate  diagrams,  it  is  seen  that  the  stress  in  any  member 
acts  in  one  direction  in  a  particular  polygon  and  in  an  oppo- 
site direction  when  repeated  in  another  polygon.  If  these 
force  polygons  are  combined,  the  line  representing  the  stress 
will  therefore  have  two  arrows  pointing  in  opposite  direc- 
tions. The  force  polygon  for  all  the  joints  of  the  truss  are 
grouped  together  into  the  stress  diagram  shown  in  Fig.  58,  i. 
If  the  diagram  is  drawn  for  the  forces  at  all  the  joints,  as  in 
this  figure,  the  last  polygon  in  the  stress  diagram  must  check 
with  one  side  on  the  line  representing  the  known  right 
reaction. 

As  the  stress  diagram  is  being  drawn,  it  is  usually  most 
convenient  to  put  the  arrows  representing  the  directions  of 
the  stresses  at  any  joint  directly  upon  the  diagram  of  the 
truss ;  as  shown  in  Fig.  58,  a.  If  this  is  done,  they  may  be 
omitted  in  the  stress  diagram. 


Art.  5. 


STRESSES  BY  GRAPHIC  RESOLUTION. 


103 


The  student  should  follow  through  the  separate  force 
polygons  in  the  stress  diagram  shown  in  Fig.  58,  i,  paying 
particular  attention  to  the  fact  that  the  forces  at  a  joint, 
whose  magnitudes  are  represented  by  a  closed  polygon  in  the 
stress  diagram,  must  act  in  the  same  direction  around  the 
polygon.  Great  care  should  be  used  in  drawing  the  truss,  and 
in  transferring  the  lines  from  the  truss  to  the  corresponding 
lines  in  the  stress  diagram  to  secure  accuracy. 

113.  Stresses  by  Graphic  Resolution.  Loads  on  Lozver 
Chord.  In  the  problem  given  in  §  112,  and  in  the  preceding 
problems  in  this  chapter,  the  loads  have  all  been  applied  at 
the  upper  chord  panel  points ;  as  it  is  the  usual  practice  to 
consider  all  the  dead  load  on  the  upper  chord.  In  some  build- 
ings, however,  the  ceiling  loads  and  other  miscellaneous  loads 
are  supported  at  the  lower  chord  panel  points ;  and  the  follow- 


X 


Stress  Diagram 

for 
Loads  on  Lower  Chord 

o*         iooo: 


FIG.  59.     STRESS  DIAGRAM: — LOADS  ON  LOWER  CHORD. 


104  STRESSES   IN   ROOF   TRUSSES.  Chap.  X. 

ing  problem  will  show  the  methods  used  in  drawing  the  stress 
diagram  for  a  truss  loaded  at  the  lower  chord  panel  points 
only. 

Problem.  It  is  required  to  find  the  stresses  due  to  ceiling 
loads  in  all  the  members  of  the  truss  shown  in  Fig.  59.  The 
trusses  are  spaced  15  ft.  apart,  each  truss  having  a  span  of  40 
ft.  and  a  rise  of  10  ft.  The  ceiling  load  is  10  Ibs.  per  sq.  ft. 

The  panel  load  P  is  equal  to   -j-   X   15  X   10  =  1000  Ibs., 

and  the  effective  reactions  are  each  equal  to  2.^/2  X  1000  = 
2500  Ibs. 

To  draw  the  stress  diagram,  lay  off  the  reactions  Rx  and 
R2  on  the  load  line,  as  shown  in  Fig.  59,  and  start  the  diagram 
with  the  forces  acting  at  the  left  end  of  the  truss.  The  stress 
diagram  is  drawn  considering  the  forces  at  the  joints  in  the 
following  order :  L0,  U1?  Llf  U2,  L2,  L3,  U3,  L2',  U/,  L/,  U/,  and 
and  L0'.  The  arrows  indicating  the  kind  of  stress  are  placed  at 
each  panel  point  of  the  truss,  as  the  portion  of  the  stress 
diagram  for  that  panel  point  is  drawn.  The  arrows,  with  the 
exception  of  those  in  the  load  line,  are  omitted  in  the  stress 
diagram.  Since  the  truss  and  the  loads  are  symmetrical  about 
the  center  line,  it  is  seen  that  the  stresses  are  all  determined 
as  soon  as  the  diagram  is  drawn  for  the  forces  up  to  and 
including  those  acting  at  L3.  It  is  therefore  unnecessary  to 
draw  the  diagram  for  the  right  half  of  the  truss. 


CHAPTER  XI. 

WIND    LOAD    STRESSES. 

This  chapter  will  treat  of  the  stresses  in  roof  trusses  due 
to  wind  loads  for  different  conditions  of  the  ends  of  the 
trusses.  The  method  for  finding  the  wind  load  stresses  in  roof 
trusses  by  graphic  resolution  will  be  shown  for  the  four  fol- 
lowing cases:  (a)  Both  Ends  of  Truss  Fixed — Reactions 
Parallel;  (b)  Both  Ends  of  Truss  Fixed — Horizontal  Com- 
ponents of  Reactions  Equal ;  (c)  Leeward  End  of  Truss  on 
Rollers ;  (d)  Windward  End  of  Truss  on  Rollers. 

ART.  i.     BOTH  ENDS  OF  TRUSS  FIXED — REACTIONS  PARALLEL. 

When  the  roof  truss  is  comparatively  flat,  it  is  usually 
customary  to  assume  that  both  reactions  are  parallel  to  the 
resultant  of  the  wind  loads.  The  wind  load  stresses  in  the 
members  of  a  roof  truss  will  be  found  by  the  method  of 
graphic  resolution,  assuming  that  the  reactions  are  parallel. 

114.  Problem.  It  is  required  to  find  the  wind  load  stresses 
in  all  the  members  of  the  truss  shown  in  Fig.  60.  The  truss 
has  a  span  of  60  ft.,  a  rise  of  15  ft.,  and  the  trusses  are  spaced 
15  ft.  apart,  center  to  center.  The  lower  chord  has  a  camber 
of  2  ft.,  and  the  normal  component  of  the  wind  is  taken  at  23 
Ibs.  per  sq.  ft.  (§  93). 

After  computing  the  loads,  the  reactions  are  found  by 
means  of  the  force  and  the  funicular  polygons  (see  §  98,  a). 
The  left  reaction  is  represented  by  Rx,  and  the  right  reaction 
by  R.2  (Fig.  60). 

105 


106 


WIND    LOAD    STRESSES. 


Chap.  XL 


The  stress  diagram  is  started  by  drawing  the  force  polygon 
for  the  forces  at  the  joint  L0  (Fig.  60).  The  stresses  in  the 
members  X-i  and  Y-i  are  unknown,  while  the  wind  load  at 

L0  and  the  left  reaction  are  known.    The  unknown  stresses  at 

p 
this   joint   are   determined   by   drawing   the    force    polygon— -=-, 

X,  I,  Y,  R!,  noting  particularly  that  the  polygon  closes,  and 
that  the  forces  act  in  the  same  direction  around  the  polygon. 


Wind    Load 
Stress  Diaqram 
O*        4000*      8000* 

FIG.  60.     ENDS  FIXED — REACTIONS  TARALLEL. 

The  lines  X-I  and  Y-i  in  the  force  polygon  represent  the 
stresses  in  the  members  X-i  and  Y-i.  By  placing  the  arrows 
on  the  members  at  the  joint  L0,  corresponding  to  the  direc- 
tions of  the  forces  in  the  force  polygon,  it  is  seen  that  X-i  is 
compression  and  that  Y-i  is  tension. 

The  force  polygon,  or  stress  diagram,  for  the  forces  at  the 
joint  H!  is  X,  X,  2,  I,  X,  the  sequence  of  the  letters  and  figures 
denoting  the  directions  of  the  forces.  The  kind  of  stress  in 


Art.  1.  ENDS   FIXED — REACTIONS   PARALLEL.  107 

the  unknown  members  X-2  and  1-2  is  determined  by  placing 
arrows  on  the  members  meeting  at  the  joint  D^.  The  stresses 
in  both  these  members  are  compression. 

The  stress  diagram  for  the  forces  at  the  joint  "L^  is  Y,  i, 
2,  3,  Y,  the  sequence  of  the  letters  and  figures  denoting  the 
directions  of  the  forces,  which  directions  are  shown  by  arrows 
placed  on  the  members  at  the  joint  Lt.  The  stresses  in  both 
2-3  and  Y~3  are  thus  found  to  be  tension. 

The  unknown  stresses  at  the  joint  U2  are  X~4  and  3-4, 
and  the  stress  diagram  for  the  forces  at  the  joint  U2  is  X,  X, 

4,  3'  2>  x- 

The  unknown  stresses  at  the  joint  L2  are  4-5  and  Y~5, 
and  the  stress  diagram  for  the  forces  at  L2  is  Y,  3,  4,  5,  Y. 

The  unknown  stresses  at  U3  are  X-6  and  5-6,  and  the  stress 
diagram  for  the  forces  at  this  joint  is  X,  X,  6,  5,  4,  X. 

At  the  joint  L3,  the  unknown  stresses  are  6-7  and  Y~7, 
and  the  stress  diagram  for  the  forces  at  this  joint  is  Y, 

5,  6,  7,  Y- 

At  the  joint  U4,  the  unknown  stresses  are  X-(6'-i')  and 
6'-7,  and  the  stress  diagram  for  the  forces  at  this  joint  is  X, 
X,  6'-!',  7,  6,  X. 

The  unknown  stress  at  the  joint  L3*  is  Y-(6'-i'),  and  the 
stress  diagram  for  the  forces  at  this  joint  is  Y,  7,  6'-i',  Y. 

The  stress  diagram  for  the  forces  at  the  joint  L0  is  Y, 
(6'-i'),  X,  Y. 

The  numerical  values  for  the  stresses  in  all  the  members 
of  the  truss  may  be  found  by  scaling  the  corresponding  lines 
in  the  stress  diagram  to  the  given  scale. 

It  is  seen  from  the  stress  diagram  that  there  are  no  stresses 
in  the  members  6/~5/,  5'-4'»  4'~3',  3'-2r,  and  2'-i'.  This  fol- 
lows, since  there  are  no  intermediate  loads  between  the  joints 
U4  and  L0',  and  the  members  X-(6'-i'),  7-6',  and  Y-(6'-i') 
form  a  triangle.  It  is  further  seen  that  it  is  necessary  to  draw 
the  stress  diagram  for  the  complete  truss,  and  that  the  mem- 
bers X-(6'-i')  and  Y-(6'-i')  must  form  a  triangle,  one  side 


108 


WIND   LOAD    STRESSES. 


Chap.  XI. 


of  which  is  the  known  reaction  R2,  which  furnishes  a  check  on 
the  work. 

ART.  2.  BOTH  ENDS  OF  TRUSS  FIXED — HORIZONTAL  COMPONENTS 
OF  REACTIONS  EQUAL. 


When  the  horizontal  component  of  the  wind  loads  is  large 
and  the  supports  are  equally  elastic,  actual  conditions  are 
most  nearly  approximated  by  taking  the  horizontal  components 
of  the  reactions  equal.  The  horizontal  component  of  the  wind 
loads  may  be  large  if  the  roof  is  steep,  or  if  a  ventilator  such 
as  shown  in  Fig.  61  is  used.  The  wind  load  stresses  in  a 
roof  truss  having  a  "monitor"  ventilator  will  now  be  found, 
assuming  that  the  reactions  have  equal  horizontal  components. 

115.  Problem.  It  is  required  to  find  the  wind  load  stresses 
in  all  the  members  of  the  truss  shown  in  Fig.  61.  The  truss 
has  a  span  of  50  ft.,  a  pitch  of  one-fourth,  and  has  a  monitor 


Wind 
Stress  Diagram 

0*        3000*      6000* 
•  t  i 

Fia.  61.     ENDS  FIXED — HOR.  COMP.  OF  REACTIONS  EQUAL. 


Art.  2.  ENDS   FIXED — HOR.    COMPS.   EQUAL.  109 

ventilator,  as  shown  in  Fig.  61.  The  trusses  are  spaced  16  ft. 
apart,  center  to  center.  The  wind  load  on  the  vertical  surface 
of  the  ventilator  is  taken  at  30  Ibs.  per  sq.  ft.,  and  the  com- 
ponent of  the  wind,  normal  to  the  roof  surface,  is  taken  at  23 
Ibs.  per  sq.  ft. 

The  wind  loads  at  the  joints  U3  and  U4  are  found  by  taking 
the  resultants  of  the  horizontal  and  inclined  loads  acting  at 
these  joints  (see  Fig.  61).  The  reactions  R!  and  R2  are 
determined  by  the  methods  explained  in  §  98,  b,  assuming  first 
that  the  reactions  are  parallel  and  finding  these  reactions  by 
means  of  the  force  and  the  funicular  polygons ;  and  then  mak- 
ing their  horizontal  components  equal. 

The  stresses  in  the  members  of  the  truss  are  found  by 
drawing  the  force  polygons  for  the  forces  at  each  joint  and 
then  combining  these  polygons,  taking  the  joints  in  the  fol- 
lowing order:  L0,  Uw  Llf  U2,  L2,  U3,  L3,  U4,  U5,  U/,  Um, 
U3',  L/,  and  L0'.  The  complete  stress  diagram  is  shown  in 
Fig.  61.  When  the  stress  diagram  has  been  drawn  for  the 
forces  up  to  those  acting  at  U3,  it  is  seen  that  there  are  three 
unknown  stresses  at  this  joint,  viz. :  5-6,  6-7,  and  X~7.  The 
stress  in  X-7  is  taken  as  zero,  and  the  load  acting  at  U4  is  held 
in  equilibrium  by  the  stresses  in  the  two  members  X-8  and 
7-8.  There  are  no  stresses  in  the  members  5'~4',  4'-3', 
3'-2',  and  2'-i'.  The  numerical  value  of  the  stresses  in  the 
members  may  be  found  by  scaling  the  corresponding  lines  in 
the  stress  diagram  to  the  given  scale.  The  kind  of  stress, 
whether  tension  or  compression,  is  given  by  the  arrows  placed 
on  the  members  of  the  truss,  as  shown  in  Fig.  61.  If  the 
arrows  act  away  from  the  center  of  the  member,  i.  e.,  toward 
the  joints,  the  stress  is  compression ;  and  if  they  act  toward 
the  center  of  the  member,  i.  e.,  away  from  the  joints,  the  stress 
is  tension. 

The  student  should  carefully  follow  through  the  construc- 
tion of  the  stress  diagram,  placing  the  arrows,  which  show  the 
directions  of  the  forces  at  the  joint,  on  the  members,  as  the 
force  polygon  for  that  joint  is  drawn. 


110  WIND   LOAD   STRESSES.  Chap.  XI. 

ART.  3.     LEEWARD  END  OF  TRUSS  ON  ROLLERS. 

If  the  span  of  the  truss  is  large,  the  change  in  its  length 
due  to  the  loads  and  to  the  temperature  variations  is  usually 
adjusted  by  placing  one  end  of  the  truss  on  rollers.  The 
stresses  in  a  roof  truss  with  its  leeward  end  supported  on 
rollers  will  now  be  determined. 

116.  Problem.  It  is  required  to  find  the  wind  load 
stresses  in  all  the  members  of  the  "Fink"  truss  shown  in  Fig. 
62,  the  leeward  end  of  the  truss  being  supported  on  rollers. 
The  span  of  the  truss  is  60  ft.,  the  rise  20  ft.,  and  the  lower 
chord  is  cambered  3  ft.  The  trusses  are  spaced  16  ft.  apart, 
and  the  normal  component  of  the  wind  is  taken  at  26  Ibs.  per 
sq.  ft. 


X 


Wind  Load 
Stress  Diagram 
*         5000*      10000* 


FIG.  62.     LEEWARD  END  OF  TRUSS  ox  ROLLERS. 

The  reactions  Rj  and  R2  (Fig.  62)  are  found  by  means  of 
the  force  and  funicular  polygons,  the  funicular  polygon  being 


Art.  3.  LEEWARD   END   ON   ROLLERS.  Ill 

started  at  the  left  point  of  support  of  the  truss  (see  §  99,  a). 
The  stresses  in  the  members  intersecting  at  the  joints  L0,  Uj, 
and  Lj  are  found  by  drawing  the  stress  diagram  for  these 
forces,  as  in  the  preceding  articles.  Referring  to  the  forces  at 
the  joint  U2,  it  is  seen  that  there  are  three  unknown  stresses 
at  this  joint,  viz.:  X~5,  4-5,  and  3-4;  and  therefore  the  stress 
diagram  cannot  be  drawn  for  this  joint.  There  are  also  three 
unknown  stresses  at  L2.  The  unknown  stresses  at  U2  may, 
however,  be  found  as  follows :  Replace  the  members  4-5  and 
5-6  by  the  auxiliary  dotted  member  connecting  the  joints  L2 
and  U3  (Fig.  62).  Now  draw  the  stress  diagram  for  the  forces 
at  the  joint  U2,  this  diagram  being  X,X,4',3,2,X.  Next  draw 
the  stress  diagram  for  the  forces  at  U3,  which  is  X,X,6,4',X. 
Now  the  line  X-6  in  the  stress  diagram  represents  the  true 
stress  in  the  member  X-6.  After  this  stress  is  determined, 
remove  the  dotted  auxiliary  member,  and  replace  it  by  the 
original  members  4-5  and  5-6.  Two  of  the  forces  (P,  and 
X-6)  at  U3  are  now  known,  and  the  stress  diagram  may  be 
drawn  for  this  joint,  this  diagram  being  X,X,6,5,X.  Four  of 
the  -forces  at  U2  are  now  known,  the  unknown  forces  being 
3-4  and  4-5.  The  stress  diagram  for  the  forces  at  this  joint  is 
X,X,5,4,3,2,X.  The  stress  diagram  for  the  forces  at  the  remain- 
ing joints  may  be  drawn  in  the  following  order:  L2,  M,  U4, 
Lo',  L0'.  There  are  no  stresses  in  the  members  6'-5',  5'~4', 
4'~3'»  3'~2/>  and  2/-i/.  The  complete  stress  diagram  for  all 
the  members  of  the  truss  is  shown  in  Fig.  62.  The  kind  of 
stress,  whether  tension  or  compression,  is  denoted  by  the 
arrows  placed  on  the  members  of  the  truss ;  and  the  numerical 
values  of  the  stresses  may  be  found  by  scaling  the  lines  in  the 
stress  diagram  to  the  given  scale. 


ART.  4.     WINDWARD  END  OF  TRUSS  ON  ROLLERS. 

Since  the  wind  may  act  from  either  of  two  opposite  direc- 
tions, it  is  necessary  to  find  the  stresses  in  the  members  of  the 
truss  when  the  rollers  are  under  the  leeward  end  of  the  truss 


112 


WIND   LOAD   STRESSES. 


Chap.  XL 


and  also  when  the  rollers  are  under  the  windward  end.  The 
stresses  in  a  roof  truss  with  its  windward  end  on  rollers  will 
now  be  found. 

117.  Problem.  It  is  required  to  find  the  wind  load 
stresses  in  all  the  members  of  the  "Camels  Back"  truss  shown 
in  Fig.  63,  the  windward  end  of  the  truss  being  supported  on 
rollers.  The  span  of  the  truss  is  60  ft.,  the  rise  14  ft.,  and  the 
trusses  are  spaced  15  ft.  apart. 


Wind  Load 
Stress  Diaqram      / 
0*         3000*       6000* /' 


FIG.  63.     WINDWARD  END  OF  TRUSS  ON  ROLLERS. 

The  normal  components  of  the  wind  loads  are  found  from 
the  diagram  based  on  Duchemin's  formula  (see  Fig.  47), 
assuming  that  P  equals  30  Ibs.  per  sq.  ft.  The  reactions  Rx 
and  R2  (Fig.  63)  are  found  by  means  of  the  force  and  funicular 
polygons  (see  §  99,  b). 

The  complete  stress  diagram  for  all  the  members  of  the 
truss  is  shown  in  Fig.  63.  The  numerical  values  of  the  stresses 
may  be  found  by  scaling  the  lines  in  the  stress  diagram  to  the 
given  scale.  The  kind  of  stress  in  each  member,  whether 
tension  or  compression,  is  shown  by  the  arrows  placed  on  the 
members  of  the  truss.  Referring  to  the  arrows  on  the  truss,  it 
is  seen  that  the  left  half  of  the  lower  chord  is  in  tension,  while 
the  right  half  is  in  compression.  It  is  also  seen  that  the  mem- 


Art.  4.  WINDWARD   END   ON   ROLLERS.  113 

bers  1-2,  3-4,  and  5-5'  are  in  tension,  while  the  corresponding 
members  i/-2/  and  3^-4'  are  in  compression;  and  further  that 
the  members  2-3  and  4-5  are  in  compression,  while  2'~3'  and 
4'-5'  are  in  tension.  Since  the  wind  may  act  from  either 
direction,  it  is  seen  that  these  members  will  be  subjected  to 
reversals  of  stress. 


CHAPTER  XII. 

STRESSES    IN    CANTILEVER    AND    UN  SYMMETRICAL 
TRUSSES— MAXIMUM   STRESSES. 

This  chapter  will  be  divided  into  two  articles.  The  first 
article  will  treat  of  the  application  of  the  method  of  graphic 
resolution  to  the  solution  of  stresses  in  cantilever  and  unsym- 
metrical  trusses,  showing  a  method  for  drawing  a  combined 
stress  diagram,  and  the  second  will  treat  of  the  determination 
of  the  maximum  stresses  in  trusses. 

ART.  i.   STRESSES  IN  CANTILEVER  AND  UNSYM METRICAL  TRUSSES. 

A  cantilever  truss  is  one  which  is  supported  at  one  end 
only,  the  other  end  being  entirely  free.  Such  a  truss  is  often 
used  to  project  over  platforms  and  entrances  to  buildings. 
The  cantilever  truss  may  be  fastened  to  the  walls  of  the  build- 
ing or  to  the  columns  supporting  the  main  trusses. 

118.  Stresses  in  a  Cantilever  Truss.  Problem.  It  is 
required  to  find  the  dead  load  stresses  in  the  cantilever  truss 
shown  in  Fig.  64.  The  span  of  the  truss  is  24  ft.  The  trusses 
are  spaced  15  ft.  apart,  center  to  center,  and  support  a  dead 
load  of  12  Ibs.  per  sq.  ft.  of  the  horizontal  projection  of  the 
roof. 

The  point  of  application  A  (Fig.  64)  and  the  line  of  action 
of  the  reaction  R2  are  known ;  while  the  point  of  application, 
only,  of  the  reaction  Rj  is  known.  The  line  of  action  of  Rl 
may  be  found  by  applying  the  principle  that,  if  a  body  is  in 
equilibrium  under  the  action  of  three  external  forces,  these 

114 


Art.  1. 


STRESSES    IN    A    CANTILEVER   TRUSS. 


115 


forces  must  all  intersect  at  a  common  point.  One  of  these 
three  forces  is  the  resultant  of  the  loads  acting  on  the  truss, 
the  other  two  being  the  two  reactions. 


8.5- 


3  I 

Stress  Diagram 
0*     1000*   ZOOO* 


FIG.  64.     STRESS  DIAGRAM — CANTILEVER  TRUSS. 

The  resultant  R  of  the  loads  acting  on  the  truss  is  found 
by  drawing  the  force  and  funicular  polygons  for  these  loads, 
as  shown  in  Fig.  64.  Now  produce  the  lines  of  action  of  R  and 
R2  until  they  intersect  at  the  point  C.  Then  the  line  of  action 
of  Rj  must  pass  through  the  points  B  and  C.  The  magnitudes 
of  these  reactions  are  found  by  drawing  the  force  triangle  for 
the  loads  and  the  two  reactions  (see  Fig.  64).  Then  ¥-7 
represents  the  magnitude  of  the  reaction  R2,  and  X~7  that  of  R±. 

The  stresses  in  the  members  of  the  truss  are  found  by 
drawing  the  stress  diagram,  starting  the  diagram  with  the 
forces  acting  at  the  point  B.  The  forces  acting  at  D  are  then 
taken,  noting  that  the  stress  in  the  member  ¥-7  is  equal  to 


116  CANTILEVER   AND   UNSYMMETRICAL   TRUSSES.    Chap.  XII. 

the  known  reaction  R2.  The  complete  stress  diagram  is  shown 
in  Fig.  64.  The  numerical  values  of  the  stresses  may  be  found 
by  scaling  the  lines  in  the  stress  diagram  to  the  given  scale. 

The  stress  diagram  might  also  have  been  drawn  without 
first  finding  the  reactions  by  starting  the  diagram  with  the 
forces  acting  at  E,  the  left  end  of  the  truss. 

Referring  to  the  truss  and  to  the  stress  diagram  (Fig.  64), 
it  is  seen  that  the  stress  in  the  member  6-7  would  be  zero  if 
the  inclination  of  the  member  Y~7  was  changed  so  that  the 
lines  of  action  of  R  and  R2  would  intersect  on  the  member 
X-2 ;  and  further,  that  the  stress  in  6-7  would  be  compression 
if  the  lines  of  action  of  R  and  R2  intersected  above  the  mem- 
ber X-2. 

119.     Unsymmetrical    Truss — Combined   Stress   Diagram. 

In  the  problems  that  have  been  given  (excepting  that  given  in 
§  113),  the  loads  have  all  been  applied  at  the  panel  points  of  the 
upper  chord  of  the  truss ;  and  separate  stress  diagrams  have  been 
drawn  for  the  dead  and  for  the  wind  loads.  This  section  will 
treat  of  the  loads  on  both  the  upper  and  lower  chords ;  and  the 
stresses  due  to  the  dead  loads  and  to  the  wind  loads  will  be  found 
by  drawing  a  single  diagram.  The  method  for  drawing  the  com- 
bined stress  diagram  will  now  be  shown  by  the  following  problem. 

Problem.  It  is 'required  to  find  the  stresses  in  the  members 
of  the  truss  shown  in  Fig.  65,  the  leeward  end  of  the  truss 
being  supported  upon  rollers.  The  span  of  the  truss  is  50  ft., 
the  rise  i6f  ft.,  and  the  adjacent  trusses  are  spaced  14  ft.  apart, 
center  to  center.  The  dead  load  is  taken  at  12  Ibs.  per  sq.  ft. 
of  horizontal  projection,  the  ceiling  load  at  10  Ibs.  per  sq.  ft. 
of  horizontal  projection,  and  the  wind  load  at  26  Ibs.  per  sq. 
ft.  of  roof  surface. 

The  resultants  of  the  dead  and  wind  loads  at  each  panel 
point  of  the  upper  chord  are  first  found.  These  resultants  are 
shown  on  the  truss  diagram  (Fig.  65).  The  reactions  Rt  and 
R2,  considering  the  right  end  of  the  truss  on  rollers,  are  found 
for  these  loads  by  means  of  the  force  and  funicular  polygons. 
These  reactions  are  shown  in  the  force  polygon,  but  are  not 
shown  on  the  diagram  of  the  truss.  The  reactions  due  to  the 


Art.  1. 


COMBINED   STRESS    DIAGRAM. 


117 


loads  on  the  lower  chord  are  then  found,  these  reactions  being 
each  numerically  equal  to  two  lower  chord  panel  loads. 


Combined 

Stress  Diagram  ^Jx^ 
0*      5000*   10000* 


FIG.  65.     UNSYMMETRICAL  TRUSS — COMBINED  STRESS  DIAGRAM. 

The  left  reaction  due  to  all  the  loads  is  then  determined  by 
rinding  the  resultant  of  the  reactions  due  to  the  dead  and 
wind  loads  on  the  upper  chord  and  to  the  ceiling  loads  on  the 
lower  chord.  This  reaction  is  represented  by  R/  (Fig.  65). 
The  right  reaction  is  determined  in  the  same  manner,  and  is 
represented  by  R2'. 

The  stresses  in  the  members  of  the  truss  are  now  found 
by  drawing  the  stress  diagram,  as  shown  in  Fig.  65.  The 
numerical  values  of  the  stresses  may  be  found  by  scaling  the 
lines  in  the  stress  diagram  to  the  given  scale. 

To  determine  which  condition  gives  the  largest  stresses,  it 
is  necessary  to  take  the  wind  as  acting  both  from  the  right 
and  from  the  left ;  and  further,  to  assume  that  the  rollers  may 
be  under  either  end  of  the  truss. 

The  above  method  of  laying  off  the  loads  in  the  load  line, 


118  MAXIMUM    STRESSES.  Chap.  XII. 

and  of  determining  the  reactions,  is  applicable  to  all  trusses 
carrying  loads  on  both  the  upper  and  lower  chords.  It  is  seen 
that  this  method  places  the  loads  in  their  proper  order  in  the 
load  line. 

ART.  2.     MAXIMUM  STRESSES. 

120.  Maximum  Stresses.  In  the  preceding  articles, 
methods  have  been  given  for  rinding  the  stresses  in  various 
types  of  trusses,  due  separately  to  the  dead  load,  to  the  snow 
load,  and  to  the  wind  load.  In  this  article,  the  maximum 
stresses  that  may  occur  in  the  members  of  a  truss  due  to  the 
combined  effect  of  the  different  loadings  will  be  found.  The 
stresses  will  first  be  found  separately  for  the  different  load- 
ings, and  then  combined  to  determine  the  maximum  stresses. 

If  the  truss  and  also  the  loads  are  symmetrical  about  the 
center  line,  each  reaction  is  equal  to  one-half  of  the  total  load 
on  the  truss  (neglecting  the  half  loads  at  the  ends).  In  this 
case,  it  is  unnecessary  to  draw  the  funicular  polygon  to  deter- 
mine the  reactions.  The  dead  load  stresses  may  then  be  found 
by  drawing  the  stress  diagram  for  the  dead  loads. 

Since  the  snow  load  is  always  taken  at  so  much  per  square 
foot  of  the  horizontal  projection  of  the  roof,  it  is  seen  that  it  is 
unnecessary  to  draw  a  stress  diagram  for  the  snow  loads ;  but 
that  the  snow  load  stresses  may  be  determined  directly  by 
proportion  from  the  known  dead  load  stresses.  The  minimum 
snow  load  in  this  work  will  be  taken  at  10  pounds  and  the 
maximum  snow  load  at  20  pounds  per  square  foot  of  the  hori- 
zontal projection  of  the  roof. 

If  the  truss  is  symmetrical  and  is  fixed  at  both  ends,  the 
wind  need  only  be  taken  as  acting  from  one  side ;  since  the 
stresses  in  the  corresponding  members  would  be  the  same  as 
if  the  wind  was  taken  from  the  other  side.  If  the  truss  has 
rollers  under  one  end,  the  wind  must  be  taken  as  acting  both 
from  the  roller  side  and  from  the  fixed  side  of  the  truss.  The 
condition  which  gives  the  larger  stresses  is  then  considered  in 
making  the  combinations  for  maximum  stresses. 


Art.  2.  MAXIMUM   STRESSES.  119 

The  dead  load  is  always  acting  upon  the  truss,  and  there- 
fore it  must  be  used  in  all  the  combinations  for  maximum 
stresses.  If  the  maximum  snow  load  is  taken,  then  the  wind 
load  will  be  neglected ;  as  it  is  improbable  that  the  maximum 
wind  and  the  maximum  snow  will  ever  occur  at  the  same 
time.  If  the  maximum  wind  is  considered,  then  the  minimum 
snow  load  will  be  used. 

From  the  above  discussion,  it  is  seen  that  the  following 
combinations  should  be  made  to  obtain  the  maximum  stresses 
in  the  members  of  a  truss  when  there  are  no  reversals  of  stress : 

(a)  Dead  load  stress  plus  maximum  snow  load  stress. 

(b)  Dead  load  stress  plus  minimum  snow  load  stress  plus 
wind  load  stress  (rollers  under  leeward  end  of  truss). 

(c)  Dead  load  stress  plus  minimum  snow  load  stress  plus 
wind  load  stress  (rollers  under  windward  end  of  truss). 

The  method  for  finding  the  maximum  stresses  in  the  mem- 
bers of  a  truss  will  now  be  shown  by  the  solution  of  a  prob- 
lem. 

121.     Problem.     It    is    required    to    find    the    maximum 
stresses  in  the  members  of  the  Fink  truss  shown  in  Fig.  66. 
The  dimensions  of  the  truss,  together  with  the  loadings  taken, 
are  shown  in  the  following  table : 
Span  of  truss  =  80  ft. 
Rise  of  truss  =  20  ft. 
Distance  between  trusses  =  16  ft. 
Dead  load  taken  at  12  Ibs.  per  sq.  ft.  hor.  proj. 
Minimum  snow  load  taken  at  10  Ibs.  per  sq.  ft.  hor.  proj. 
Maximum  snow  load  taken  at  20  Ibs.  per  sq.  ft.  hor.  proj. 
Wind  load  taken  at  23  Ibs.  per  sq.  ft.  of  roof  surface. 
Since  one-half  of  the  upper  chord  is  divided  by  the  web 
members    into    eight   equal    parts,   preliminary    computations 
give  the  following  panel  loads : 

Dead  panel  load 960  Ibs. 

Minimum  snow  panel  load 800  Ibs. 

Maximum  snow  panel  load 1600  Ibs. 

Wind  panel  load 2060  Ibs. 

Wind  loads  at  end  and  apex 1030  Ibs. 


120 


MAXIMUM    STRESSES. 


Chap.  XII. 


Since  the  dead  loads  are  symmetrical  about  the  center  line, 
the  effective  dead  load  reactions  are  each  equal  to  one-half  of 
the  total  dead  loac*  (neglecting  the  half  loads  at  the  ends  of 
the  truss).  The  dead  load  stresses  are  found  by  drawing  the 
stress  diagram,  as  shown  in  Fig.  66,  b. 


Ow  2000*4000*6000* 


Dead   Load 
Stress  Diaqram 


(c) 

Wind    Load 
Stress  Diagrams 
0*  8000*4000*6000* 

I  I  I  I 

FIG.  66.     STRESS  DIAGRAMS — FINK  TRUSS    (No   CAMBER) 

The  minimum  and  maximum  snow  load  stresses  are 
obtained  from  the  dead  load  stresses  without  drawing  another 
stress  diagram.  The  ratio  of  the  minimum  snow  load  to  the 


Art.  2.  MAXIMUM   STRESSES.  121 

dead  load  is  as  5  is  to  6,  and  that  of  the  maximum  snow  load 
to  the  dead  load  is  as  5  is  to  3.  The  minimum  and  maximum 
snow  load  stresses  are  obtained  by  multiplying  the  correspond- 
ing dead  load  stresses  by  these  ratios. 

The  wind  load  reactions,  considering  the  leeward  end  of 
the  truss  on  rollers,  are  found  by  means  of  the  force  and 
funicular  polygons.  These  reactions  are  represented  by  Rj 
and  R2  (Fig.  66,  c).  The  wind  load  stress  .diagram  is  shown 
in  Fig.  66,  c.  The  apparent  ambiguity  at  some  of  the  joints 
is  overcome  by  substituting  the  dotted  members  in  the  truss 
diagram  and  the  corresponding  ones  in  the  stress  diagram,  as 
is  shown  in  Fig.  66,  a  and  Fig.  66,  c  (see  §  116). 

The  wind  load  reactions,  considering  the  windward  end  of 
the  truss  on  rollers,  are  represented  by  R/  and  R2'  (Fig.  66, 
c).  The  wind  load  stress  diagram  for  this  case  is  also  shown 
in  Fig.  66,  c.  It  is  seen  that  the  stresses  in  all  the  tipper  chord 
and  web  members  for  this  case  are  the  same,  and  that  the 
stresses  in  the  lower  chord  members  are  smaller  than  they 
were  when  the  rollers  were  considered  under  the  leeward  end. 

The  stresses  in  the  members  of  this  truss  due  to  the  differ- 
ent loadings,  together  with  the  maximum  stresses,  are  shown 
in  tabular  form  in  the  upper  half  of  the  table  given  in  Fig.  68. 
It  is  seen  that,  for  this  particular  truss,  there  are  no  reversals 
of  stress  in  any  of  the  members. 

122.  Problem.  It  is  required  to  find  the  maximum 
stresses  in  the  members  of  the  truss  shown  in  Fig.  67,  the  lower 
chord  being  cambered  two  feet.  The  other  dimensions  of  the 
truss,  together  with  the  dead,  snow,  and  wind  loads,  are  the  same 
as  for  the  problem  given  in  the  preceding  section. 

The  minimum  and  maximum  snow  load  stresses  are  obtained 
by  proportion  from  the  dead  load  stresses  (see  §  121). 

The  dead  load  stress  diagram  is  shown  in  Fig.  67,  b. 

The  wind  load  stress  diagram,  considering  the  leeward  end 
of  the  truss  on  rollers,  is  shown  by  the  full  lines  in  Fig.  67,  c. 
The  wind  load  stress  diagram,  considering  the  windward  end 
of  the  truss  on  rollers,  is  shown  by  the  broken  lines  in  Fig. 
67,  c. 


122 


MAXIMUM   STRESSES. 


Chap.  XII. 


The  stresses  in  the  members  of  the  truss  due  to  the  dif- 
ferent loadings,  together  with  the  maximum  stresses,  are 
shown  in  the  lower  half  of  the  table  given  in  Fig.  68.  It  is 
seen  that,  for  this  truss,  there  are  no  reversals  of  stress  in  any 
of  the  members. 

By  comparing  the  stresses  given  in  Fig.  68,  it  is  seen  that 
even  a  small  camber  in  the  lower  chord  increases  the  stresses 


(c) 

Wind  Load 
Stress  Diagrams 
0*          5000*     10000 


Fio  67.     STRESS  DIAGRAMS — FINK  TRUSS    (WITH     CAMBER). 

in  most  of  the  members  a  considerable  amount.  A  small 
camber,  however,  improves  the  appearance  of  the  truss,  and 
also  increases  the  clearance  in  the  building.  If  no  camber  is 


MAXIMUM   STRESSES. 


123 


STRESSES  IN  A  FINK 


Truss 
Member 

Dead 
Load 
Stress 

Snow  Load 
Stress 

Wind  Load  Stress 

Max- 
imum 
Stress 

Rollers 
_eeward 

Rollers 
Windward 

Min. 

Max- 

X-l 

-16100 

-  1  3400 

-26800 

-20600 

-20600 

-50100 

X-2 

-  1  5700 

-13100 

-26200 

-20600 

-20600 

-49400 

X-5 

-  1  5300 

-  12700 

-25500 

-  20600 

-20600 

-48600 

X-6 

-14800 

-  12300 

-  24700 

-  20600 

-  20600 

-47700 

X-9 

-14400 

-  12000 

-24000 

-  20600 

-  20600 

-47000 

x-io 

-  14000 

-1  1700 

-23300 

-20600 

-20600 

-  46300 

X-13 

-13600 

-  I  1300 

-22700 

-  20600 

-20600 

-45500 

X-14 

-13100 

-  10900 

-21800 

-20600 

-20600 

-44600 

X-04M1) 

-  10300 

-  10300 

Y-l 

-hi  4400 

+  12000 

+  24000 

+  25400 

+  1  8000 

+  51800 

Y-3 

-1-13500 

+  1  1300 

+  22500 

+  23000 

-1-  15700 

+  47800 

Y-7 

+  1  1  600 

+   9700 

+  19300 

+  18400 

+  II  100 

+  39700 

Y-15 

+   7800 

+   6500 

+  13000 

+    9200 

+     1900 

+  23500 

Y-dS-l1) 

+    9200 

+     1900 

1-8,5-6,9-10,13-14 

-     800 

-      700 

-    1300 

-    2100 

-    2100 

-   3600 

2-3,4-5,10-11,12-13 

+     900 

-1-      800 

+     1500 

+    2300 

+    2300 

+    4000 

3-4,11-12 

-    1700 

-    1400 

-    2800 

-   4100 

-   4100 

-    7200 

4-7,8-11 

+    1400 

+     1200 

+    2300 

+    4600 

+    4600 

+    7ZOO 

6-7,8-9 

t    2800 

+    2300 

+    4700 

+-    6900 

+    6900 

+  12  000 

7-8 

-   3400 

-    2800 

-     5700 

-    8200 

-    8200 

-  14400 

8-15 

+   3800 

+    3200 

+    6300 

+    9ZOO 

+    9ZOO 

+  16200 

12-15 

+   5700 

'+   4300 

+    9500 

+  13800 

+  13800 

+  24300 

14-15 

+   6700 

+    5600 

+  1  1200 

+  16100 

+  16100 

+  28400 

STRESSES  IN  A  FINK 


IS  (With  Cam! 


x-i 

-19300 

-16100 

-32200 

-26?00 

-  24500 

-61600 

X-2 

-18800 

-15700 

-31400 

-26ZOO 

-24500 

-60700 

X-5 

-18400 

-15300 

-30700 

-26200 

-24500 

-59900 

X-6 

-18000 

-15000 

-30000 

-26200 

-24500 

-  59200 

X-9 

-17600 

-14700 

-29300 

-26200 

-24500 

-58500 

X-IO 

-17200 

-14300 

-28600 

-26200 

-  24500 

-  57700 

X-13 

-16700 

-13900 

-27800 

-26200 

-24500 

-  56800 

X-14 

-16300 

-13600 

-27200 

-26200 

-  24500 

-56100 

X-(I4'-I') 

-12300 

-  1  0700 

Y-l 

-hi  7300 

+  14400 

+  28800 

+30400 

+  21600 

+  62100 

Y-3 

+I6ZOO 

+13500 

+  27000 

+27700 

+18800 

+  57400 

Y-7 

+13900 

+  1  1600 

+  23ZOO 

+22200 

+13300 

+  47700 

Y-15 

+  8600 

+    7ZOO 

+  14300 

+10200 

+    Z  000 

+  26000 

Y-r/'-n 

+  1  1  100 

+    2200 

1-2,5-6,9-10,13-14 

-      800 

-      700 

-    1300 

-2100 

-    2100 

-    3600 

2-3,4-5,10-11,12-13 

+    MOO 

+      900 

+    1800 

+   2800 

+    2800 

+  4800 

3-4.M-I2 

-    1700 

-    1400 

-    Z800 

-  4100 

-   4100 

-    7200 

4-7,8-11 

+    2300 

+    1900 

+    3800 

+    5500 

+    5500 

+    9700 

6-7,8-9 

+    3400 

+    2800 

+    5700 

+    8300 

+    8300 

+  14500 

7-8 

-   3400 

-    Z800 

-    5100 

-    8200 

-    8200 

-  144-00 

8-15 

+    5600 

+  4700 

+   9300 

+12300 

+  11300 

+  22600 

12-15 

+    7900 

+    6600 

+13200 

+  17800 

+16800 

+  32300 

14-15 

+   9000 

+    7500 

+15000 

+20600 

+  19600 

+  31100 

l5-(l4'-8'} 

+    1200 

+      200 

FIG.  68.     TABLE  OF  STRESSES — MAXIMUM   STRESSES. 


124  MAXIMUM   STRESSES.  Chap.  XII. 

used  and  the  span  is  long,  the  lower  chord  has  the  appearance 
of  sagging. 

123.  Maximum  and  Minimum  Stresses.  By  the  terms 
"maximum  and  minimum  stresses"  are  meant  the  greatest 
ranges  of  stress  that  may  occur  in  any  member  due  to  the, 
different  loadings.  If  there  is  no  reversal  of  stress  in  any 
member,  the  maximum  stress  is  the  greatest  numerical  stress, 
and  the  minimum  is  the  smallest  numerical  stress  that  may 
ever  occur  in  any  member.  If  there  is  a  reversal  of  stress,  the 
maximum  is  the  greatest  numerical  stress,  and  the  minimum 
is  the  greatest  numerical  stress  of  an  opposite  kind  that  may 
ever  occur. 

In  finding  maximum  and  minimum  stresses,  it  must  be 
borne  in  mind  that  the  dead  load  is  always  acting,  and  that 
there  is  no  reversal  of  stress  in  any  member  unless  the  wind 
load  stress,  -or  stress  due  to  another  condition  of  loading,  in 
that  member  is  greater  in  amount  and  is  of  an  opposite  kind 
to  that  caused  by  the  dead  load. 

In  designing,  there  is  no  need  of  finding  minimum  stresses 
unless  there  are  reversals  of  stress ;  since  if  there  are  no  rever- 
sals, the  members  must  be  designed  for  their  maximum 
stresses,  while  if  there  are  reversals,  they  must  be  designed  for 
each  kind  of  stress. 

If  both  the  maximum  and  minimum  stresses  are  required, 
the  following  combinations  should  be  made : 

(a)  Dead  load  stress  alone. 

(b)  Dead  load  stress  plus  wind  load  stress   (rollers  lee- 
ward). 

(c)  Dead  load  stress  plus  wind  load  stress  (rollers  wind- 
ward). 

(d)  Dead  load  stress  plus  maximum  snow  load  stress. 

(e)  Dead    load    stress   plus    minimum    snow    load    stress 
plus  wind  load  stress  (rollers  leeward). 

(f)  Dead  load  stress  plus  minimum  snow  load  stress  plus 
wind  load  stress  (rollers  windward). 


CHAPTER  XIII. 

.       COUNTERBBACING. 

The  use  of  counterbracing  adds  considerably  to  the  work 
required  to  find  the  maximum  stresses  in  a  truss ;  since  the 
same  diagonals  are  not  always  stressed.  In  many  cases,  how- 
ever, its  use  is  more  economical  than  to  design  the  member  for 
a  reversal  of  stress.  There  are  two  methods  which  may  be 
used  to  determine  the  stresses  in  a  truss  with  counterbracing, 
viz. :  (a)  by  the  use  of  separate  stress  diagrams,  and  (b)  by 
the  use  of  combined  stress  diagrams.  The  former  method 
may  be  used  to  advantage  when  several  different  combinations 
must  be  made  to  determine  the  maximum  stresses,  while  the 
latter  is  useful  when  but  few  combinations  are  required. 

The  determination  of  stresses  in  trusses  with  counterbrac- 
ing will  now  be  taken  up  in  three  articles,  as  follows:  Art.  i, 
Definitions  and  Notation ;  Art.  2,  Stresses  in  Trusses  with 
Counterbracing — Separate  Stress  Diagrams;  and  Art.  3, 
Stresses  in  Trusses  with  Counterbracing — Combined  Stress 
Diagrams. 

ART.  i.     DEFINITIONS  AND  NOTATION. 

124.  Definitions.  The  triangle  is  the  only  geometrical 
figure  which  is  incapable  of  any  change  in  shape  without  a 
change  in  the  length  of  one  or  more  of  its  sides.  The  triangle 
is  therefore  the  elementary  form  of  truss;  and  the  trusses  in 
common  use  consist  of  a  series  of  triangles  so  arranged  as  to 
form  a  rigid  body. 

A  polygonal  figure  which  is  composed  of  more  than  three 

125 


126 


COUNTERBRACING. 


Chap.  XIII. 


sides  and  which  is  free  to  turn  at  its  joints  may  be  distorted 
without  changing  the  length  of  any  of  its  sides.  For  example, 
take  the  quadrilateral  frame  shown  in  Fig.  69,  a,  the  frame 
being  acted  upon  by  an  external  force.  This  figure  may  be 
distorted  without  changing  the  length  of  any  of  its  sides,  as 
is  shown  by  the  dotted  lines  in  the  figure. 


B 


B 


B 


,rA' 


D  C 

(a) 


CD  C 

(b)  (c) 

FIG.  69. 


Now  suppose  a  diagonal,  connecting  the  points  B  and  D, 
is  added  to  the  frame  ABCD,  as  shown  in  Fig.  69,  b.  The 
figure  is  then  composed  of  two  triangles,  and  any  force  acting 
towards  the  right  will  tend  to  distort  the  frame,  and  will  pro- 
duce tension  in  the  diagonal  BD.  The  tension  produced  in  this 
diagonal  will  prevent  any  distortion  of  the  frame.  Now  sup- 
pose that  the  external  force  acts  towards  the  left,  as  shown  in 
Fig.  69,  c.  The  distortion  of  the  frame  will  be  prevented  by 
the  diagonal  BD,  which  will  be  in  compression. 

It  is  thus  seen  that  the  diagonal  BD,  which  is  capable  of 
resisting  both  tension  and  compression,  will  prevent  any  dis- 
tortion in  the  frame  ABCD.  The  same  is  true  of  the  diagonal 
AC.  If  the  diagonal  member  is  capable  of  resisting  tension 
only,  or  compression  only;  then  two  diagonals  will  be  re- 
quired to  prevent  distortion,  as  shown  in  Fig.  69,  d.  If  two 
diagonals  are  used,  and  both  are  designed  to  take  the  same 
kind  of  stress,  it  is  evident  that  only  one  acts  at  a  time. 
Referring  to  Fig.  69,  d,  if  both  the  diagonals  are  rods  and 
therefore  can  resist  tension  only,  it  is  seen  that  BD  is  in 
tension  when  the  external  force  acts  towards  the  right  and 
that  there  is  no  stress  in  AC.  It  is  further  seen  that  AC  is  in 


Art.  1. 


DEFINITIONS    AND    NOTATION. 


127 


tension  when  the  force  acts  towards  the  left  and  that  there  is 
no  stress  in  BD.  If  two  diagonals  are  used,  and  each  can 
resist  both  compression  and  tension,  the  problem  is  indeter- 
minate by  static  methods.  This  case  will  not  be  considered 
in  this  work. 

Let  the  quadrilateral  ABCD  (Fig.  70)  be  one  panel  of  a  Pratt 
truss     loaded     with     dead 
load,  as  shown  in  the  fig- 
ure.    In  this  type  of  truss, 
the  intermediate  diagonals 

are  tension  members.    The       f  D       C        D' 

members     AC     and     A'C,        |  I 

which  are  in  tension  under 

the  dead  load,  are  called  main  diagonals  or  main  braces;  while 
the  members  BD  and  BD',  which  are  not  stressed  by  the  dead 
load,  are  called  counterbraces  or  counters.  The  counters  may  be 
stressed  either  under  the  action  of  wind  loads  or  of  unsymmetrical 
loads. 

125.  Counterbracing.  The  two  tension  or  two  compres- 
sion diagonals  in  the  same  panel  cannot  act  at  the  same  time 
unless  they  are  subjected  to  initial  stress.  A  truss  will  not  be 
in  equilibrium  under  the  action  of  initial  stress  unless  external 
forces  are  applied,  or  unless  the  initial  stress  is  held  in  equi- 
librium by  the  resisting  moment  of  some  of  the  members  of 
the  truss.  Since  initial  stress  will  not  be  considered  in  this 
work,  it  is  evident  that  if  the  diagonals  can  resist  only  one 
kind  of  stress,  but  one  diagonal  in  each  panel  will  act  at  a 
time. 

The  method  for  determining  whether  the  main  member  or 
the  counter  is  under  stress  due  to  any  system  of  loading, 
together  with  the  magnitude  of  the  resulting  stress,  will  now 
be  given.  Referring  to  Fig.  70,  and  noting  the  fact  that  the 
dead  load  is  always  acting  upon  the  truss,  it  may  be  readily 
shown  that  there  will  be  a  tensile  stress  in  the  diagonal  AC 
when  the  dead  load  alone  is  acting.  The  kind  of  stress  in  this 
member  may  be  determined  by  drawing  the  stress  diagram,  or 


128  COUNTEKBRACING.  Chap.  XIII. 

by  a  method  which  will  be  explained  in  the  following  article. 
Now  suppose  that  the  snow  load  covers  the  right  half  of  the 
truss  only,  or  that  the  wind  is  acting  from  the  right.  Either 
of  these  conditions  will  tend  to  cause  a  compressive  stress  in 
the  diagonal  A'C.  If  this  compression  is  less  than  the  dead 
load  tension  in  the  member,  then  the  tension  already  in  the 
member  will  be  reduced  by  an  amount  equal  to  the  magnitude 
of  the  compressive  stress.  The  resultant  of  these  stresses  will 
be  the  actual  stress  in  the  member  due  to  the  combined  loads. 
If  this  compression  is  exactly  equal  to  the  dead  load  tension, 
the  resulting  stress  in  the  member  is  zero.  Now  suppose  that 
the  compression  caused  by  the  unsymmetrical  loading  is 
greater  than  the  dead  load  tension.  In  this  case,  the  member 
A'C  can  only  take  enough  compression  to  neutralize  the  dead 
load  tension.  If  more  than  this  amount  of  compression  is 
thrown  into  A'C,  the  member  will  tend  to  be  distorted,  and 
the  counter  BD'  will  be  thrown  into  tension  to  resist  this 
distortion. 

If  the  chords  are  parallel  and  the  diagonals  have  the  same 
inclination,  as  in  the  truss  shown  in  Fig.  70,  the  magnitude  of 
the  tension  in  the  counter  BD'  is  equal  to  the  difference 
between  the  stress  in  A'C  caused  by  the  wind  or  unsym- 
metrical load  and  that  caused  by  the  dead  load.  If  the  chords 
are  not  parallel,  i.  e.,  if  the  main  member  and  the  counter  in 
any  panel  have  different  inclinations,  the  stress  in  the  counter 
is  equal  to  the  difference  in  shears  in  the  panel  resolved  in  the 
direction  of  the  counter. 

126.  Notation.  When  counterbracing  is  used,  the  system 
of  notation  must  be  slightly  modified ;  since  either  diagonal  in 
a  panel  may  be  stressed.  A  convenient  system  of  notation  is 
shown  in  Fig.  71.  In  this  system,  one  diagonal  in  each  panel 
is  shown  as  a  dotted  line.  The  diagonals  shown  as  full  lines 
are  designated  by  the  figures  2-3  and  4-5 ;  while  those  shown 
as  dotted  lines  are  designated  by  the  same  figures  accented, 
2'-3'  and  4'-5'. 

To    illustrate    this    system    of    notation,    if   the    diagonals 


Art. 


SEPARATE    STRESS    DIAGRAMS. 


129 


stressed  are  2'-$'  and  4-5,  then  the  upper  chord  members  are 
X-i,  X-3',  X-5,  and  X-6;  the 
lower  chord  members  are  Y-i, 
Y-2',  Y-4,  and  Y-6;  and  the  web 
members  are  1-3',  2'-$',  2'-$,  4-5, 
and  4-6.  If  the  dead  load  alone  is 
acting,  the  diagonals  2-3  and  4-5 
are  stressed,  and  the  web  mem- 
bers are  1-2,  2-3,  3-5,  4-5,  and  4-6. 


FIG.  71.     NOTATION. 


ART.  2.     STRESSES  IN  TRUSSES  WITH  COUNTERBRACING — 
SEPARATE  STRESS  DIAGRAMS. 

127.  Determination  of  Stresses  in  Trusses  with  Counter- 
bracing — Separate  Stress  Diagrams.  In  the  explanations  that 
follow,  the  diagonals  will  be  assumed  to  be  tension  members, 
as  this  is  the  more  common  case;  although  the  same  general 
principles  will  apply  if  they  are  compression  members.  The 
method  for  determining  the  maximum  stresses  in  trusses  with 
counterbracing,  when  separate  stress  diagrams  are  used,  in- 
volves the  following  steps : 

(i).  Construct  the  dead  and  snow  load  stress  diagrams, 
assuming  that  the  diagonals  all  slope  in  the  same  direction.  If 
the  truss  is  symmetrical  about  its  center  line,  it  will  only  be 
necessary  to  construct  the  diagrams  for  one-half  of  the  truss. 
The  snow  load  stresses  may  be  found  by  direct  proportion  from 
the  dead  load  stresses  without  the  use  of  another  diagram. 

(2).  From  these  diagrams,  determine  in  which  panels,  if 
any,  the  diagonals  will  be  subjected  to  compression,  and  draw 
in  the  second  diagonals  in  these  panels.  Revise  the  stress 
diagrams  to  include  the  added  diagonals.  The  revised  diagrams 
will  now  contain  the  actual  stresses  in  all  the  members  due  to 
vertical  loads.  The  main  members  (those  stressed  by  the  dead 
load)  are  represented  by  full  lines,  and  the  counters  by  dotted 
lines. 

(3).     Construct  the  wind  load  stress  diagrams,  using  those 


130  COUNTEBBRACING.  Chap.  XIII. 

diagonals  which  have  been  found  to  be  in  tension  due  to  the 
dead  load.  If  the  truss  is  symmetrical,  it  will  only  be  necessary 
to  consider  the  wind  as  acting  from  one  direction ;  while  if  the 
truss  is  unsymmetrical,  it  will  be  necessary  to  consider  the 
wind  as  acting  from  both  directions. 

(4).  From  the  wind  load  stress  diagram,  determine  which 
diagonals,  if  any,  will  be  in  compression.  Draw  counters  in 
these  panels.  Revise  the  stress  diagrams  to  include  the  added 
diagonals. 

(5).  From  the  stress  diagrams,  determine  the  stresses  due 
to  the  different  loadings,  combine  these  stresses  to  determine 
which  diagonals  are  stressed,  and  then  find  the  maximum 
stresses  in  all  the  members  of  the  truss.  In  making  any  par- 
ticular combination  to  determine  the  maximum  stress  in  any 
member,  it  is  necessary  to  first  find  which  diagonals  are  acting 
at  that  time,  and  then  to  combine  the  stresses  found  in  that 
particular  member  when  these  diagonals  are  acting.  In  mak- 
ing the  combinations  for  maximum  stresses,  it  is  necessary  to 
consider,  not  only  the  member  itself,  but  also  the  corresponding 
member  on  the  other  side  of  the  center  of  the  truss. 

The  following  combinations  will  be  considered  in  this  work 
in  determining  maximum  stresses  in  roof  trusses. 

(a)  Dead  load  plus  maximum  snow  load. 

(b)  Dead  load  plus  minimum  snow  load  plus  wind  load 
(wind  acting  from  either  direction). 

(c)  Dead  load  plus  wind  load   (wind  acting  from  either 
direction). 

The  method  outlined  above  will  now  be  explained  by  the 
solution  of  two  problems.  The  first  problem  will  be  to  deter- 
mine the  maximum  stresses  in  a  truss  with  parallel  chords,  and 
the  second  to  determine  the  maximum  stresses  in  a  truss  with 
non-parallel  chords. 

128.  Problem  i.  Truss  With  Parallel  Chords.  It  is 
required  to  find  the  maximum  stresses  in  all  the  members  of 
the  Pratt  truss  with  counterbracing  shown  in  Fig.  72.  The 
span  of  the  truss  is  40  ft. ;  the  height,  7.5  ft. ;  and  the  trusses 
are  spaced  15  ft.  apart.  The  dead  load  will  be  taken  at  10  Ibs. 


Art. 


SEPARATE    STRESS    DIAGRAMS. 


131 


per  sq.  ft.  of  hor.  proj. ;  the  minimum  snow  load,  at  10  Ibs.  per 
sq.  ft.  of  hor.  proj. ;  the  maximum  snow  load,  at  20  Ibs.  per  sq. 
ft.  of  hor.  proj. ;  and  the  component  of  the  wind  normal  to  the 
roof  surface,  at  27  Ibs.  per  sq.  ft.  of  roof  surface.  The  wind 
load  reactions  will  be  assumed  parallel  to  the  resultant  of  all 
the  wind  loads. 


FIG.  72.     STRESS  DIAGRAMS — TRUSS  WITH  PARALLEL  CHORDS. 

The  dead  load  stress  diagram  (see  Fig.  72)  is  first  drawn, 
assuming  that  all  the  diagonals  slope  in  the  same  direction,  i.  e., 
that  2-3  and  4'~5'  are  acting.  From  this  diagram  it  is  found 
that  4'-5',  if  acting,  would  be  in  compression;  therefore,  the 
other  diagonal,  4-5,  in  the  same  panel  is  acting  due  to  the  dead 
load.  The  stress  diagram  is  now  revised  to  include  the  diagonal 
4-5,  also  the  diagonal  2/~3/.  The  dead  load  stress  diagram  for 
the  entire  truss  is  shown  in  Fig.  72.  The  dead  load  stresses 
when  the  different  diagonals  are  acting  are  shown  in  the  table 
in  Fig.  73. 

The  minimum  and  maximum  snow  load  stresses  are  deter- 


132 


COUNTERBRACING. 


Chap.  XIII. 


mined  by  direct  proportion  from  the  dead  load  stresses  without 
constructing  any  new  diagrams,  and  are  shown  in  Fig.  73. 


Truss 
Mem- 
ber 

Dead 
Load 
Stress 

Snow  Load 
Stress 

Wind 
Load 
Stress 

Ma  xi  - 
mumv 
Stress 

Min. 

Max. 

X-l 

-  3750 

-  3750 

-  7500 

-  2070 

-1  1250* 

rx-31 

-  3000 

-  3000 

-  6000 

-  3160 

— 

1  X-3 

-  4000 

-  4000 

-  8000 

-  2110 

-12000* 

rx-5 

-  4000 

-  4000 

-  8000 

-  21  10 

-12000 

(X-51 

-  3000 

-  3000 

-  6000 

-  1060 

X-6 

-  3750 

-  3750 

-  7500 

-  1320 

-1  1250 

Y-l 

-1-  3000 

4  3000 

4  6000 

4  2580 

-T  9000* 

/  Y-21 

-h  4000 

4  4000 

-I-  6000 

4  1530 



lY-Z 

4  3000 

4  3000 

+  6000 

+  2580 

+  9000* 

/Y-41 

-I-  4000 

4  4000 

+•  8000 

4  1530 



1  Y-4 

4  3000 

4  3000 

+  6000 

+  480 

4  9000 

Y-6 

+  3000 

-1-  3000 

4  6000 

+  480 

4  9000 

f  1-3' 

4   750 

4   750 

+  1500 

-  790 

-   40* 

1  1-2 

0 

0 

0 

0 

0 

f  2'-3' 

-  1250 

-  1250 

-  2500 

+  1320 

4   70* 

1  2-3 

4  1350 

4  1250 

4  2500 

-  1320 

4  3750 

3-5 

-  1500 

-  1500 

-  3000 

0 

-  4500* 

l2'-5 

-  750 

-  750 

-  1500 

-  790 



(4'-5' 

-  1250 

-  1250 

-  2500 

-  1320 



I  4-5 

4  1250 

4  1250 

4  2500 

+  1^20 

4-  38EO* 

(4-6 

0 

0 

0 

0 

0 

1  5'-6 

+   750 

+   750 

-1-  1500 

-  790 



FIG.  73.  TABLE  OF  STRESSES — TRUSS  WITH  PARALLEL  CHORDS. 

The  wind  load  stress  diagram  for  the  wind  acting  from  the 
left  is  also  shown  in  Fig.  72.  The  diagram  is  first  drawn  using 
the  diagonals  which  are  found  to  be  in  tension  for  dead 
load,  viz. :  2-3  and  4-5.  The  stresses  due  to  the  wind  load 
are  shown  in  Fig.  73.  By  comparing  the  dead  and  wind  load 
stresses,  it  is  seen  that  the  wind  tends  to  cause  compression 
in  2-3,  and  further  that  the  wind  load  compression  in  this 
member  is  greater  than  the  dead  load  tension ;  therefore,  the 
counter  2^-3'  will  act.  The  stress  diagram  (Fig.  72)  is  now 
revised  to  include  the  counter  2'~3',  also  the  counter  4'-5' ; 
since  the  latter  member  will  be  stressed  when  the  wind  acts 
from  the  right.  Since  all  the  members  which  may  ever  act 
are  included  in  the  stress  diagram  and  the  truss  is  symmetri- 


-Art-  -•  SEPARATE    STRESS    DIAGRAMS.  133 

cal,  it  is  evident  that  it  is  unnecessary  to  construct  the  wind 
load  stress  diagram  for  the  wind  acting  from  the  right.  The 
wind  load  stresses  when  the  different  diagonals  are  acting  are 
given  in  Fig.  73.  The  maximum  stresses  in  all  the  mem- 
bers are  determined  by  making  the  combinations  indicated 
in  §  127,  and  are  shown  in  the  last  column  of  Fig.  73.  From 
this  table,  it  is  seen  that  the  counters  2'~3'  and  4'~5'  are 
required.  These  counters  are  shown  by  dotted  lines  in  the 
truss  and  stress  diagrams.  In  determining  the  maximum 
stresses,  it  is  seen  that  not  only  the  member  itself,  but  also 
the  corresponding  member  on  the  other  side  of  the  truss  must 
be  considered.  The  stresses  shown  with  a  star  after  them  in 
Fig.  73  are  maximum  stresses. 

For  a  truss  with  parallel  chords,  it  is  evident  that,  if  the 
diagonals  2-3  and  2'~3'  act  separately,  the  stresses  in  them  are 
numerically  equal  but  have  opposite  signs,  i.  e.,  if  one  is  ten- 
sion, the  other  will  be  compression.  It  is  therefore  seen  that 
it  would  not  be  necessary  to  actually  use  the  member  2/~3/ 
in  the  stress  diagrams. 

129.  Problem  2.  Truss  With  Non-parallel  Chords.  It  is 
required  to  find  the  maximum  stresses  in  all  the  members  of 
the  truss  with  counterbracing  shown  in  Fig.  74.  The  span  of 
the  truss  is  100  ft. ;  the  total  height,  37.5  ft. ;  the  height  of 
the  vertical  sides,  12.5  ft. ;  the  pitch  of  the  roof,  one-fourth ; 
and  the  trusses  are  spaced  15  ft.  apart.  The  panel  points  of 
the  lower  chord  lay  on  the  circumference  of  a  circle  of  106.25 
ft.  radius.  The  dead  load  is  taken  at  10  Ibs.  per  sq.  ft.  of  hor. 
proj.;  the  minimum  snow  load,  at  10  Ibs.  per  sq.  ft.  of  hor. 
proj.;  the  maximum  snow  load,  at  20  Ibs.  per  sq.  ft.  of  hor. 
proj.;  the  wind  on  the  vertical  sides  of  the  truss,  at  30  Ibs. 
per  sq.  ft.  of  surface ;  and  the  component  of  the  wind  normal 
to  the  roof  surface,  at  23  Ibs.  per  sq.  ft.  of  roof  surface.  The 
wind  load  reactions  are  assumed  parallel  to  the  resultant  of 
all  the  wind  loads. 

The  dead  load  stress  diagram  for  one-half  of  the  truss  is 
shown  in  Fig.  74.  In  constructing  this  diagram,  it  was  first 
assumed  that  all  the  diagonals  sloped  downward  toward  the 


134 


COUNTERBRACING. 


Chap.  XIII. 


right.  From  the  stress  diagram,  it  is  seen  that  the  dead  load 
tends  to  cause  compression  in  5'-6'  and  7'-8/ ;  therefore,  5-6 
and  7-8  are  stressed  by  the  dead  load.  The  stress  diagram 
was  then  revised  to  include  these  members. 


Dead  Load 
Stress  Diagram 


FIG.    74.     NON-PARALLEL    CHORDS — DEAD    LOAD    STRESS  DIAGRAM. 

The  minimum  and  maximum  snow  load  stresses  are  found  by 
proportion  from  the  dead  load  stresses  without  constructing  addi-- 
tional  diagrams. 

The  wind  load  stress  diagram,  assuming  the  wind  to  act 
upon  the  left  side  of  the  truss,  is  shown  in  Fig.  75.  This  dia- 


Art. 


SEPARATE   STRESS   DIAGRAMS. 


135 


gram  was  constructed  by  first  using  the  diagonals  which  are 
in  tension  due  to  the  dead  load. 


\ 


Wind     Load 
Stress  Diaqram 


0*      3000**  6000*   9000*  12000* 


FIG.    75.     NON-PARALLEL    CHORDS — WIND    LOAD    STRESS  DIAGRAM. 

The  table  of  stresses  is  shown  in  Fig.  76,  the  stresses  being 
determined  from  the  diagrams  in  Fig.  74  and  Fig.  75.  This 
table  is  constructed  as  follows :  First  record  the  stresses  in 
the  diagonals,  starting  with  the  member  1-2.  The  dead  and 
wind  load  stresses  are  obtained  from  the  stress  diagrams,  and 
the  minimum  and  maximum  snow  load  stresses  are  obtained 
by  direct  proportion  from  the  dead  load  stresses.  The  dead 


136  COUNTERBRACING.  Chap.  XIII. 

and  wind  loads  produce  the  same  kind  of  stress  in  the  diag- 
onals 1-2,  3-4,  5-6,  and  7-8 ;  therefore  no  counters  are  required 
in  the  left  half  of  the  truss  when  the  wind  acts  towards  the 
right.  The  dead  load  stress  in  9-10  is  +  3500,  and  the  wind 
load  stress  in  that  member  is  — 2400;  therefore  no  counter 
is  required  in  that  panel.  The  dead  load  stress  in  11-12  is 
+  300,  and  the  wind  load  stress  in  that  member  is  —  4300. 
Since  the  member  11-12  can  only  take  enough  wind  load  com- 
pression to  neutralize  the  dead  load  tension  already  in  it,  it  is 
seen  that  the  counter  u'-i2'  will  be  thrown  into  action. 
Revise  the  wind  load  stress  diagram  to  include  the  counter 
n/-i2',  as  shown  in  Fig.  75.  Also  revise  the  dead  load  stress 
diagram  to  include  the  corresponding  counter  $'-6'  (if  not 
already  included).  Record  the  member  n'-i2'  in  the  table, 
and  tabulate  the  dead,  snow,  and  wind  load  stresses.  Also 
record  the  stresses  in  the  diagonals  13-14  and  15-16.  It  is 
seen  that  the  dead  and  wind  loads  produce  the  same  kind  of 
stress  in  these  members,  therefore  no  counters  are  required  in 
these  panels.  It  is  thus  seen  that  n'-i2'  is  the  only  counter 
required  in  the  truss  when  the  wind  acts  towards  the  right. 

Likewise,  tabulate  the  stresses  in  the  verticals.  The 
stresses  in  the  verticals  adjacent  to  the  diagonal  11-12  should 
be  considered,  both  when  the  main  member  11-12  and  when 
the  counter  11'— 12'  act,  i.  e.,  when  11-12  acts,  the  verticals 
are  10-11  and  12-13,  and  when  n'-i2/  acts,  the  verticals  are 
10-12'  and  n/-i3. 

Also,  tabulate  the  stresses  in  the  upper  and  lower  chord 
members.  The  stresses  in  the  chord  members  in  the  same 
panel  as  1 1-12  should  be  considered,  both  when  11-12  and 
when  ii'-i2'  act. 

Combine  the  stresses  to  determine  the  maximum  stress  in 
each  member,  as  shown  in  Fig.  76.  In  making  the  combina- 
tions, it  must  be  borne  in  mind  that,  to  get  the  stress  in  any 
member,  it  is  necessary  to  first  determine  which  diagonals  are 
acting  for  each  combination,  and  to  combine  the  stresses  in 
that  particular  member  when  these  diagonals  are  acting.  The 
combinations  considered  are  those  indicated  in  §  127.  Since 


Art.  2. 


SEPARATE    STRESS    DIAGRAMS. 


137 


Truss 
Mem- 
ber 


Dead 

Load 

Stress 


Snow  Load 
Stress 


Mirv 


Max- 


Wind 

Load 

Stress 


Max- 
imum 
Stress 


l-Z  ' 

3-4 

5-6 

7-8 

9-10 

ii'-iz'l 

13-14 

15-16  . 

A- 1 

Z-3 

4-6 

5-8 

7-9 

10-11 

IO-IZ' 

IZ-13 

ir-13. 

14-15 

A- 16 

X-Z 

X-4 

X-6 

X-8 

X-IO 

X-1Z 

X-IZ' 

X-13 

X-15J 

Y-3 
Y-5 

Y-7 

Y-9 

Y-ll 

Y-tl1 

Y-14 

Y-16  J 


7ZOO 
3600 

300 
3500 
3500 

300 

-  ZOO 
+    3600 
4-    7ZOO 

-  7600 

-  5500 

-  1900 

-  ZIOO 
4-     1000 

-  ZIOO 

-  1900 

-  1900 

-  1700 

-  5500 

-  7600 

-  7100 

-  10300 

-  10500 

-  10300 

-  10300 

-  10500 

-  10300 

-  10300 
-7100 

0 

6700 
9300 
7600 
7600 
9300 
9500 
6700 

0 


7ZOO 
3600 

300 
3500 
3500 

300 

-  ZOO 
4-  3600 
4-  7ZOO 

-  7600 

-  5500 

-  1900 

-  ZIOO 
4-  1000 

-  ZIOO 

-  1900 

-  1900 

-  1700 

-  5500 

-  7600 

-  7100 

-  10300 

-  10500 
-10300 

-  10300 
-10500 

-  10300 

-  10300 

-  7100 

0 

4-  6700 
4-  9300 
4-  7600 
4-  7600 
4-  9300 
4-  9500 
4-  6700 

0 


4-  14400 
4-  7ZOO 
4-  600 
t  7000 
4-  7000 
4-  600 

-  400 
4-  7ZOO 
4-  14400 

-  I5ZOO 

-  11000 

-  3800 

-  4ZOO 
4-  ZOOO 

-  4ZOO 

-  3600 

-  3800 

-  3400 

-  I  1000 

-  I5ZOO 

-  I4ZOO 
-Z0600 

-  Z1000 

-  Z0600 

-  Z0600 
-Z1000 

-  Z0600 

-  Z0600 

-  I4ZOO 

0 

4-  13400 
4-  18600 
4-  I5ZOO 
4-  I5ZOO 
t  18600 
4-  19000 
4-  13400 

0 


4-  9700 
4-  1ZOO 
4-  8400 
4-  13700 

-  Z400 

-  4300 
+    3ZOO 
4-   4ZOO 
4-    4500 

-  13100 

-  6600 

-  5400 

-  10600 


4-   500 

4-  Z700 

0 

0 

-  3400 

-  4ZOO 

-  4ZOO 

-  14000 

-  17400 
-19800 

-  17500 

-  10900 

-  8400 

-  10900 

-  8400 

-  4500 
4-  6300 
4-  15100 
4-  1  I  300 
4-  5ZOO 
4-  5ZOO 
4-  4100 
4-  1900 

-  1700 

-  6300 


4-Z4IOO* 
4-  10800 
4-    9000* 
4-Z0700* 
4-  10500 
4-      900 
4-    3000* 
4-  1  1400* 
4-ZI600 
-Z8300* 

-  17600* 

-  9ZOO* 

-  14800* 
4-    3000* 

-  6300 

-  5700 

-  5700 

-  6800 

-  16500 

-  ZZ800 
-Z8ZOO* 
-38000* 
-40800* 
-38100* 
-31500 
-31500 
-31500 

-  30900 
-ZI300 
4-    6300* 
4-  Z8500* 
4-Z9900* 
4-ZZ800* 
4-  ZZ600 
4-  Z7900 
4-Z0900 
4-ZOIOO 

-  6300* 


FIG.  76.     TABLE  OF  STRESSES — TRUSS  WITH  COUNTERBRACING. 


138  COUNTERBRACING.  Chap.  XIII. 

the  wind  is  taken  as  acting  towards  the  right,  only,  it  is  nec- 
essary to  consider  the  member  itself  and  also  the  correspond- 
ing member  on  the  other  side  of  the  center  line.  The  maximum 
stresses  are  indicated  by  stars  in  Fig.  76. 

Referring  to  the  table  of  stresses,  it  is  seen  that  the  maxi- 
mum stress  in  1-2  is  given  by  the  combination  of  dead,  mini- 
mum snow,  and  wind  loads;  that  the  maximum  stress  in  7-9 
is  given  by  the  combination  of  dead  and  maximum  snow  loads ; 
and  that  the  maximum  stress  in  the  counter  ii'-i2'  is  given 
by  the  combination  of  dead  and  wind  loads.  The  attention  of 
the  student  is  called  to  the  stress  Y-II',  which  is  the  stress 
in  the  lower  chord  when  the  counter  n'-i2'  is  acting.  Unless 
care  is  taken  to  determine  which  diagonal  is  acting,  the  stu- 
dent is  liable  to  make  the  mistake  of  combining  the  dead  and 
maximum  snow  load  stresses,  which  would  seem  to  give  a 
stress  of  +28500  in  Y-II'.  However,  this  stress  can  never 
occur,  as  the  main  diagonal  11-12  is  acting  for  this  combina- 
tion. When  the  dead,  minimum  snow,  and  wind  loads  are  on 
the  truss,  the  counter  n'-i2'  is  acting,  and  the  stress  in 
Y-II'  is  then  +20900,  which  is  the  maximum  stress  in  that 
member  when  the  counter  is  acting.  When  the  main  diagonal 
11-12  is  acting,  the  combination  of  dead  and  maximum  snow 
loads  gives  a  stress  of  +27900  in  Y-n.  However,  the  maxi- 
mum stress  occurs  in  the  corresponding  member  Y~5  when 
the  dead,  minimum  snow,  and  wind  loads  are  acting,  and  is 
+  29900.  It  is  seen  that  the  stress  in  Y-II  is  also  +29900 
when  the  wind  acts  towards  the  left.  Referring  to  Y-i  and 
Y-i6,  it  is  seen  that  the  stresses  in  these  members  are  +  6300 
and  —  6300,  respectively. 

If  the  direction  of  the  wind  is  changed  and  is  made  to  act 
towards  the  left,  it  is  seen  that  Y-i  and  Y-i6  are  subjected  to 
reversals  of  stress.  It  is  further  seen  that  the  counter  5'-6' 
is  then  thrown  into  action.  The  diagonals  5'-6'  and  n/-i2' 
are  the  only  counters  required,  and  Y-i  and  Y-i6  are  the 
only  members  which  have  reversals  of  stress. 

The  maximum  stresses  are  shown  on  the  truss  diagram  in 
Fig.  76. 


Art-  3-  COMBINED   STRESS   DIAGRAM.  139 

In  this  problem,  it  has  been  shown  that  the  only  counter 
required  when  the  wind  acts  towards  the  right  is  n'-i2/.  In 
some  trusses,  it  might  happen  that  when  the  wind  acts  towards 
the  right  a  counter  is  required  in  some  panel  on  the  left  of  the 
center  and  also  in  another  panel  (not  the  corresponding  one)  on 
the  right  of  the  center  of  the  truss.  In  this  case,  it  would  be  more 
convenient  and  would  give  less  cause  for  errors  in  making  the 
combinations  for  maximum  stresses  if  the  dead  load  stress  dia- 
gram was  drawn  for  the  entire  truss.  The  diagram  should  also 
be  revised  to  include  both  the  main  diagonal  and  the  counter  in 
each  panel  in  which  a  counter  is  required  when  the  wind  acts 
towards  the  right. 

ART.  3.     STRESSES  IN  TRUSSES  WITH  COUNTERBRACING — 
COMBINED  STRESS  DIAGRAM. 

130.  Determination  of  Stresses  in  Trusses  With  Counter- 
bracing  —  Combined  Stress  Diagram.  In  determining  the 
maximum  stresses  in  a  truss  with  counterbracing  by  means  of 
the  combined  stress  diagram,  it  is  necessary  to  first  find  which 
diagonal  in  each  panel  is  stressed  by  the  combined  loadings. 
The  stress  diagram  is  then  constructed,  using  only  those  diag- 
onals which  are  found  to  be  stressed.  Two  methods  will  now 
be  given  for  finding  which  diagonals  are  stressed.  The  first 
method  may  be  used  to  advantage  for  trusses  with  parallel 
chords,  and  the  second,  for  trusses  with  non-parallel  chords. 

(i)  Trusses  with  Parallel  Chords.  If  the  truss  has  parallel 
chords,  the  simplest  method  for  determining  which  diagonal 
in  each  panel  is  stressed  consists  of  an  application  of  the  con- 
dition of  equilibrium  that  S  V  =  o  to  the  external  forces  on 
one  side  of  a  section  and  the  members  cut  by  the  section. 
This  method  will  now  be  explained  by  means  of  a  problem. 

It  is  required  to  find  which  diagonals  are  stressed  in  the 
truss  loaded  as  shown  in  Fig.  77.  The  load  line,  together  with 
the  reactions,  is  shown  in  Fig.  77,  b.  To  determine  whether  2-3 
or  2'-3'  is  acting,  cut  the  members  X~3,  2-3,  2^3',  and  Y-2  by 


140 


COUNTERBHACING. 


Chap.  XIII. 


the  section  m-m,  and  apply  the  condition  that  2  V  =  o  to  the 
forces  acting  upon  the  shaded  portion  of  the  truss.     Since  the 

chords  are  horizontal 
members  and  can 
no   vertical    force, 


5000     1000       1000 


-r 

I 

I 

IRi 
I 
I 
"t 


.1 

(b 


FIG.  77. 


resist 
it    is 

seen  that  the  vertical 
component  of  the  stress 
in  the  diagonal  2-3,  or 
in  the  diagonal  2^-3', 
must  be  equal  and  oppo- 
site in  direction  to  the  re- 
sultant of  the  external 
forces  to  the  left  of  the 
section  m-m.  Now  the 
represented  in  the  load 
therefore,  the 


resultant  of  these  external  forces  is 
line  (Fig.  77,  b)  by  EB,  and  acts  downward; 
vertical  component  of  the  stress  in  the  diagonal  must  act  upward 
for  equilibrium.  Referring  to  Fig.  77,  it  is  seen  that  if  a  mem- 
ber is  in  tension,  the  force  it  exerts  upon  the  shaded  portion  of 
the  truss  acts  away  from  that  portion ;  and  if  it  is  in  compres- 
sion, the  force  it  exerts  acts  toward  the  shaded  portion.  There- 
fore, if  the  diagonals  are  tension  members,  as  they  are  in  the 
Pratt  truss  shown  in  Fig.  77,  the  member  2-3  will  act  in  tension. 
In  like  manner,  it  may  be  shown  that  the  diagonal  4-5  is  also  in 
action.  From  the  above  discussion,  it  is  seen  that  by  observing 
whether  the  resultant  of  the  external  forces  to  the  left  of  any 
panel  acts  upward  or  downward  and  knowing  whether  the  diag- 
onals are  tension  or  compression  members,  it  may  be  determined 
at  once  which  diagonal  in  any  panel  is  stressed.  The  method 
described  above  is  called  the  method  of  shears. 

The  stresses  may  now  be  found  by  constructing  the  stress 
diagrams  for  the  combined  loads,  using  only  those  diagonals 
which  have  been  found  to  be  stressed.  It  is  seen  that  this 
method  requires  considerable  work  if  several  combinations 
must  be  made  to  determine  the  maximum  stresses;  as  it  is 
necessary  to  determine  for  each  combination  which  diagonals  are 
stressed,  and  then  to  draw  a  stress  diagram  for  each  combination. 


Art.  3. 


COMBINED    STRESS    DIAGRAM. 


141 


(2)  Trusses  with  Non-parallel  Chords.  If  the  chords  of  the 
truss  are  not  parallel,  the  condition  of  equilibrium  that  2  M  =  o 
may  be  used  to  determine  which  diagonals  are  stressed.  The 
application  of  this  method  will  now  be  shown  by  means  of  a 
problem. 

It  is  required  to  determine  which  diagonals  are  stressed  in 
the  truss  loaded  with  dead  and  wind  loads,  as  shown  in 
Fig.  78. 


FIG.  78. 

The  reactions  (assumed  to  be  parallel)  are  first  determined 
by  means  of  the  force  and  funicular  polygons,  as  shown  in 
Fig.  78.  To  determine  whether  2-3  or  2,'-$'  is  in  action  due 
to  the  combined  loads,  cut  these  members,  together  with  the 
chord  members  in  the  same  panel,  by  the  section  m-m  ;  and 
consider  the  members  cut  and  the  external  forces  to  the  left  of 
the  section.  Prolong  the  chord  members  X~3  and  Y-2  until 
they  intersect  at  P,  and  take  this  point  as  the  center  of 
moments.  Now  from  the  condition  that  2  M  =  o,  the  moment 
of  the  stresses  in  the  members  cut  by  the  section  must  balance 
the  moment  of  the  external  forces  to  the  left  of  the  section 
m-m.  But  the  moment  of  each  chord  stress  is  zero,  since  its 
line  of  action  passes  through  the  center  of  moments;  there- 
fore, the  moment  of  the  stress  in  the  diagonal  2-3,  or  the 
diagonal  2'-3',  must  balance  the  moment  of  the  external  forces 
to  the  left  of  the  section  m-m.  The  next  step  is  to  determine 


142  COUNTERBRACING.  Chap.  XIII. 

in  which  direction  the  moment  of  the  external  forces  to  the 
left  of  m-m  tends  to  produce  rotation.  •  Now  the  resultant  of 
these  external  forces  is  represented  in  the  force  polygon  by 
GC,  acting  in  the  direction  from  G  towards  C;  and  its  line 
of  action  is  through  the  intersection  of  the  strings  oc  and  og. 
Although  the  intersection  of  these  ctrings  falls  outside  of  the 
limits  of  the  drawing,  it  is  evident  that  the  moment  of  the 
resultant  of  the  external  forces  is  clockwise;  therefore,  the 
moment  of  the  stress  in  the  diagonal  acting  at  this  time  must 
be  counter-clockwise.  In  the  truss  shown  in  Fig.  78,  the  diag- 
onals are  tension  members;  therefore,  the  diagonal  2-3  is 
stressed. 

In  like  manner,  by  using  the  section  n-n  and  taking  the 
center  of  moments  at  P',  it  is  found  that  the  moment  of  the 
resultant  GE  of  the  external  forces  to  the  left  of  the  section 
is  counter-clockwise ;  hence  the  moment  of  the  stresses  in  the 
diagonal  6-7,  or  6'~7',  is  clockwise.  The  diagonal  6-7  is 
therefore  stressed. 

To  determine  whether  4-5  or  4'~5'  is  stressed,  cut  the  mem- 
bers by  the  section  p-p.  Instead  of  using  the  method  of 
moments  for  this  case,  which  would  necessitate  first  finding 
the  kind  of  stress  in  X~5  or  ¥-4,  the  method  of  shears, 
described  in  §  130  (i),  will  be  used.  From  the  force  polygon, 
it  is  seen  that  the  vertical  component  of  the  resultant  GD  of 
the  external  forces  to  the  left  of  the  section  p-p  acts  down- 
ward; therefore,  the  vertical  component  of  the  stress  in  the 
diagonal  must  act  upward.  The  diagonal  4-5  is  therefore 
stressed. 

The  stresses  may  now  be  found  by  constructing  the  stress 
diagrams  for  the  combined  loads,  using  only  those  diagonals 
which  have  been  found  to  be  in  action  due  to  the  combined 
loading. 

It  is  seen  that  this  method  ^pay  be  used  to  advantage  if 
only  a  single  combination  is  required  to  determine  the  maxi- 
mum stresses. 


CHAPTER  XIV. 

THREE-HINGED  ARCH. 


131.  Definition.  An  arch  is  a  structure  which  has  inclined 
reactions  for  vertical  loads.  The  only  type  of  arch  which  will 
be  considered  here,  and  the  one  commonly  used  for  long  span 
roof  trusses,  is  the  three-hinged  arch.  A  three-hinged  arch  is 
composed  of  two  simple  beams  or  trusses,  hinged  together  at 
the  crown,  and  also  hinged  at  their  points  of  support.  This  is 
the  only  form  of  arch  construction  which  is  statically  determi- 
nate. An  example  of  a  three-hinged  arch  is  shown  in  Fig. 
79,  a.  This  structure  is  composed  of  two  simple  trusses, 
hinged  together  at  the  crown  C  and  also  hinged  at  the  supports 
A  and  B. 

The  reactions  may  be  obtained  by  applying  the  principles 
which  have  already  been  explained.  The  method  of  finding 
these  reactions  may  best  be  explained  by  first  determining  the 
reactions  due  to  a  single  load. 


FIG.   79.     THREE-HINGED  ARCH — REACTIONS  FOR  SINGLE   LOAD. 


132.  Reactions  Due  to  a  Single  Load.  It  is  required  to 
find  the  reactions  in  the  three-hinged  arch  shown  in  Fig.  79,  a, 
due  to  the  single  load  P. 

143 


144  THREE-HINGED   ARCH.  Chap.  XIV. 

This  arch  is  composed  of  the  two  segments  AC  and  BC, 
and  is  hinged  at  the  points  A,  B,  and  C.  The  load  P  is  sup- 
ported by  the  segment  AC,  the  other  segment  being  unloaded. 
There  are  only  two  forces  acting  upon  the  segment  BC,  the  force 
exerted  by  the  segment  AC  against  the  segment  BC  (acting 
downward),  and  the  reaction  R2  of  the  segment  BC  against 
its  support.  Since  this  segment  is  held  in  equilibrium  by 
these  two  forces,  they  must  have  the  same  line  of  action  and 
must  act  in  opposite  directions. 

Now  consider  the  segment  AC.  This  segment  is  held  in 
equilibrium  by  three  forces,  viz.:  the  reaction  Rj  at  A,  the 
load  P,  and  the  force  (acting  upward)  exerted  by  the  segment 
BC  against  the  segment  AC.  This  last  force  is  equal  in  mag- 
nitude, but  acts  in  an  opposite  direction  to  the  force  exerted 
by  the  segment  AC  against  the  segment  BC;  and  is  also  equal 
to  the  reaction  R2.  Since  the  segment  is  held  in  equilibrium 
by  these  forces,  they  must  intersect  at  a  common  point.  This 
point  is  determined  by  prolonging  the  line  of  action  of  the 
reaction,  which  acts  through  the  points  B  and  C,  until  it  inter- 
sects the  line  of  action  of  the  force  P  at  D  (Fig.  79,  a).  The 
reactions  at  C  will  neutralize  each  other,  and  the  three-hinged 
arch,  taken  as  a  whole,  is  in  equilibrium  under  the  action  of 
the  three  forces,  R1?  P,  and  R2.  The  lines  of  action  of  the  two 
reactions  R^  and  R2  being  known,  the  magnitudes  of  these 
reactions  may  be  determined  by  drawing  their  force  polygon. 
This  polygon  is  shown  in  Fig.  79,  b,  the  reactions  being 
represented  by  Rx  and  R2. 

The  reaction  at  either  support  due  to  any  number  of  loads 
may  be  found  by  first  determining  that  due  to  each  load  sepa- 
rately, and  then  combining  these  separate  reactions. 

The  method  for  finding  the  reactions  and  stresses  due  to  a 
number  of  vertical  loads  will  now  be  shown. 

133.  Reactions  and  Stresses  for  Dead  Load.  It  is  required 
to  find  the  dead  load  reactions  and  stresses  for  the  three- 
hinged  arch,  loaded  as  shown  in  Fig.  80,  a.  The  span  of  the 
arch  is  125  ft.,  and  its  rise  is  50  ft.  Both  the  segments  AC 
and  BC  are  alike,  and  are  symmetrically  loaded. 


DEAD   LOAD    STRESSES. 


145 


To  find  the  reactions,  lay  off  the  load  line,  as  shown  in 
Fig.  80,  b,  choose  any  pole  O,  and  draw  the  funicular  poly- 
gons, one  for  each  segment,  as  shown  in  the  figure.  The  ver- 
tical reactions  at  supports  A  and  B  and  at  the  hinge  C  are 


Dead   Load  R\ 

Stress  Diagram  \ 

o*     10000*  aoooo*          \  N 


(b) 


FIG.    80.     STRESS   DIAGRAM   FOR  A   THREE-HINGED  ARCH. 

found  by  drawing  the  rays  from  the  pole  O  parallel  to  the 
closing  strings  of  these  funicular  polygons.  These  reactions 
are  represented  by  Ra,  Rb,  and  (R0  +  R'J,  respectively. 
Since  the  truss  and  the  loads  are  symmetrical  about  the  center 
line,  half  of  the  load  at  the  crown,  or  Rc,  will  be  transferred 


146  THREE-HINGED   ARCH.  Chap.  XIV. 

to  the  left  support,  and  the  other  half,  or  Rct  will  be  trans- 
ferred to  the  right  support.  To  determine  the  reaction  at  A 
due  to  all  the  loads,  consider  first  the  reaction  at  this  point 
due  to  the  single  load  Rc  acting  at  C.  Since  there  can  be  no 
resisting  moment  at  the  hinge  C,  it  follows  that  the  reaction 
at  A  caused  by  this  load  must  also  pass  through  the  hinge  C, 
which  gives  its  line  of  action.  Now  the  vertical  component 
of  this  reaction  is  equal  to  Rc,  and  its  line  of  action  is  AC; 
therefore,  its  magnitude  is  given  by  the  line  R/  (Fig.  80,  b), 
drawn  from  the  point  E  parallel  to  the  line  joining  the  points 
A  and  C.  The  total  reaction  at  A  due  to  all  the  loads  is  now 
determined  by  combining  the  vertical  reaction  Ra  with  the 
reaction  R/.  This  total  reaction  is  represented  by  Rt  (Fig. 
80,  a  and  Fig.  80,  b).  The  reaction  at  B  is  determined  in  like 
manner,  and  is  represented  by  R2.  Taking  the  equilibrant  of 
the  reaction  Rj  and  of  the  resultant  of  the  loads  acting  upon 
the  segment  AC,  it  is  seen  that  the  reaction  at  C  is  horizontal 
and  acts  towards  the  left.  This  reaction  is  represented  by  FY 
(Fig.  80,  b).  In  like  manner,  it  may  be  shown  that  the  reac- 
tion at  C,  due  to  the  loads  on  the  segment  BC,  is  horizontal 
and  acts  toward  the  right,  and  is  represented  by  YF. 

The  reactions  having  been  determined,  the  stresses  in  the 
members  of  the  three-hinged  arch  may  be  found  by  drawing 
the  stress  diagram,  starting  the  diagram  with  the  forces  acting 
at  A.  Since  the  truss  and  the  loads  are  symmetrical  about  the 
center  hinge,  it  is  only  necessary  to  draw  the  stress  diagram 
for  one  segment.  The  stress  diagram  for  the  segment  AC  is 
shown  in  Fig.  80,  b.  The  magnitudes  of  the  stresses  may  be 
determined  from  the  stress  diagram,  and  the  kind  of  stress  is 
indicated  by  the  arrows  placed  on  the  members  of  the  truss, 
as  shown  in  Fig.  80,  a. 

134.  Wind  Load  Stresses  for  Windward  Segment  of 
Truss.  It  is  required  to  find  the  wind  load  stresses  in  the 
windward  segment  of  the  three-hinged  arch,  loaded  as  shown 
in  Fig.  81,  a. 

To  find  the  reactions  for  this  segment,  consider  it  as  a  sim- 
ple truss  supported  at  the  hinges.  Construct  the  force  polygon 


WIND   LOAD    STRESSES. 


147 


(Fig.  81,  b)  for  the  wind  loads,  choose  any  pole  O,  and  draw 
the  funicular  polygon,  as  shown  in  Fig.  81,  a.  The  reactions 
are  then  determined  by  drawing  the  ray  through  the  pole  O 
parallel  to  the  closing  string  of  the  funicular  polygon.  These 
\ 


X 


Wind    Load 
Stress  Diaqram 

For 

Windward  Side 
0*        10000*   30000* 


STRESS  DIAGRAM  FOR  A  THREE-HINGED  ARCH. 


reactions  are  parallel  to  the  resultant  of  the  wind  loads,  and 
are  represented  by  Ra  and  Rc  (Fig.  81,  b).  Now  consider  the 
three-hinged  arch  as  a  whole.  Since  there  are  no  loads  on  the 
segment  BC,  for  equilibrium,  the  right  reaction  R2,  acting  at 
B,  must  also  pass  through  the  center  hinge  C.  The  reaction 
Rc,  which  was  found  by  considering  the  segment  AC  as  a 
simple  truss,  will  now  be  resolved  into  the  two  reactions  R2 
and  R/  (Fig.  81,  b),  drawn  parallel  to  R2  and  R/  (Fig.  81,  a), 
respectively.  The  reaction  Rt  is  found  by  drawing  the  closing 
line  of  the  force  polygon.  The  arch  is  held  in  equilibrium  by 


148  THREE-HINGED    ARCH.  Chap. 

the  three  following  forces,  viz. :  the  resultant  R  of  all  the  wind 
loads,  the  reaction  Rx  at  A,  and  the  reaction  R2  at  B.  Since 
the  line  of  action  of  R2  is  known,  that  of  Rx  may  be  determined 
by  prolonging  R  and  R2  until  they  intersect,  and  connecting 
this  point  of  intersection  with  A.  In  this  problem,  the  point 
of  intersection  of  R  and  R2  falls  outside  the  limits  of  the  drawing. 

The  reactions  may  also  be  determined,  as  follows:  Since 
there  are  no  loads  on  the  right  segment,  the  line  of  action  of 
R2  must  pass  through  the  hinges  B  and  C.  Therefore,  choose 
any  pole  O,  and,  starting  at  A,  the  only  known  point  on  the 
left  reaction  Rx,  draw  the  funicular  polygon  for  the  given  truss 
and  loads,  closing  on  the  line  of  action  of  R2.  The  dividing 
ray,  drawn  parallel  to  the  closing  string,  will  then  determine 
the  two  reactions. 

The  latter  method  is  somewhat  simpler  than  the  one 
shown  in  Fig.  81. 

The  reactions  having  been  determined,  the  stresses  in  the 
windward  segment  may  be  found  by  drawing  the  stress  dia- 
gram, starting  the  diagram  with  the  forces  acting  at  A.  This 
diagram  is  shown  in  Fig.  81,  b.  The  magnitudes  of  the  stresses 
may  be  determined  from  the  stress  diagram,  and  the  kind  of 
stress  is  indicated  by  the  arrows  placed  on  the  members  of  the 
segment  AC. 

To  determine  the  maximum  and  minimum  stresses,  it  is 
also  necessary  to  find  the  stresses  in  the  members  of  the  lee- 
ward segment  BC,  and  these  stresses  will  be  determined  in  the 
following  section. 

135.  Wind  Load  Stresses  for  Leeward  Segment  of  Truss. 
It  is  required  to  find  the  wind  load  stresses  in  the  leeward 
segment  of  the  three-hinged  arch  shown  in  Fig  82,  a,  the  arch 
and  the  loads  being  the  same  shown  as  in  Fig.  81,  a. 

To  facilitate  a  comparison  of  stresses  in  the  windward  and 
leeward  segments,  the  stresses  will  be  found  in  the  same 
segment  AC  as  in  the  preceding  section. 

The  wind  load  reactions  are  the  same  as  those  found  in 
§  134,  and  are  represented  in  magnitude  in  Fig.  82,  b  and  in 
line  of  action  in  Fig.  82,  a.  The  wind  load  stresses  are  found 


WIND   LOAD   STRESSES. 


149 


by  drawing  the  stress  diagram,  starting  the  diagram  with  the 
forces  acting  at  A.  This  diagram  is  shown  in  Fig.  82,  b.  The 
magnitude  of  the  stresses  may  be  determined  from  the  stress 
diagram,  and  the  kind  of  stress  is  indicated  by  the  arrows 
placed  on  the  segment  AC. 


X  ™z 


Wind   Load 
Stress  Diagram 

For 
Leeward  Side 

o*      10000*   ^oooo* 


(b) 


1.2 

Fio.   82.     STRESS   DIAGRAM   FOR  A  THREE-HINGED  ARCH. 

By  comparing  the  wind  load  stress  diagrams  and  also  the 
kind  of  stress  indicated  by  the  arrows  (see  Fig.  81  and  Fig. 
82),  it  is  seen  that  there  are  many  reversals  of  stress. 

The  maximum  stresses  may  be  determined  by  making  the 
combinations  indicated  in  §  127. 


CHAPTER  XV. 

STEESSES  IN  A   TRANSVERSE  BENT  OF  A  BUILDING. 

136.  Construction  of  a  Transverse   Bent.      It   has   been 
assumed  in   the  preceding   discussion   of   roof  trusses   that  the 
trusses  were  supported  upon  walls  or  pilasters.  However,  in  many 
types  of  mill  building  construction,  the  trusses  are  supported  by 
columns,  to  which  they  are  rigidly  connected,  thus   forming  a 
transverse  bent.     The  columns  carry  not  only  the  roof  trusses 
and  the  loads  on  them,  but  may  also  support  the  side  covering 
and  resist  the  pressure  of  the  wind  against  the  side  of  the  build- 
ing.    In  such  buildings,  the  trusses  are  usually  riveted  to  the 
columns  and  are  braced  by  members  called  knee-braces.     The 
side  covering  is  fastened  to  longitudinal  girts,  which  are  con- 
nected to  the  columns.     Additional  rigidity  is  secured  by  means 
of  a  system  of  wind  bracing.    This  bracing  may  be  placed  in  the 
planes  of  the  sides  of  the  building  and  in  the  planes  of  the  upper 
and  lower  chords  of  the  trusses. 

The  intermediate  transverse  bents  support  a  full  panel  load; 
while  those  at  the  end  carry  but  half  a  panel  load,  and  are  some- 
times made  of  lighter  construction.  The  end  bent  may  be  built 
by  running  end  posts  up  to  the  rafters,  or  the  same  construction 
may  be  used  as  for  the  intermediate  transverse  bents.  The  latter 
method  is  preferable  when  an  extension  in  the  length  of  the 
building  is  contemplated. 

137.  Condition  of  Ends  of  Columns.     The  stresses  in  a 
transverse  bent  depend  to  a  considerable  extent  upon  the  condi- 
tion of  the  ends  of  the  columns.     Several  assumptions  may  be 
made,  although  it  is  difficult,  if  not  impossible,  to  exactly  realize 
any  of  the  assumed  conditions.     The  columns  may  be  taken  as 
(i)  hinged  at  the  top  and  base,  (2)  hinged  at  the  top  and  fixed  at 
the  base,  or  (3)  rigidly  fixed  at  the  top  and  base. 

150 


CONDITION  OF  ENDS  OF  COLUMNS.  151 

(1)  Columns  Hinged  at  Top  and  Base.     If  the  columns 
merely  rest  upon  masonry  piers,  or  if  no  effectual  attempt  is 
made  to  fix  them  at  the  base  by  embedding  the  columns  in  con- 
crete or  by  fastening  them  with  anchor  bolts,  the  columns  should 
be  taken  as  hinged  at  the  top  and  base.    The  common  assumption, 
and  the  one  which  will  be  made  in  this  text,  is  that  the  horizontal 
components  of  the  reactions  due  to  the  wind  are  each  equal  to 
one-half  of  the  horizontal  component  of  the  total  external  wind 
load  acting  upon  the  structure.    The  vertical  components  may  be 
found  by  the  method  of  moments,  or  by  means  of  the  graphic 
construction  shown  in  §  139. 

The  maximum  bending  moment  in  the  column  is  at  the  foot 
of  the  knee-brace  of  the  leeward  column,  and  is  equal  to  the 
horizontal  component  of  the  reaction  multiplied  by  the  distance 
from  the  foot  of  the  knee-brace  to  the  foot  of  the  column,  i.  e.,  = 
H2d  (see  Fig.  83). 

(2)  Columns  Hinged  at  Top  and  Fixed  at  Base.     If  the 
deflections  at  B,  the  foot  of  the  knee-brace  (Fig.  83),  and  at  C, 
the  top  of  the  column,  are  assumed  to  be  equal,  it  may  be  shown 
that  the  vertical  components  of  the  reactions  are  the  same  as  if 
the  columns  were  hinged  at  D,  the  point 

of  contra-flexure.     The  distance  y  from 

the  base  of  the  column  to  the  point  of     /_ 

contra-flexure  depends  upon  the  relative  ' 

lengths  of  h  and  d.     It  may  be  shown 

that  y  varies  from  f d  when  d  =  Jh,  to  |d 

when  d  =  h.     As  soon  as  the  position 

of  the  point  of  contra-flexure  has  been 

found,    the    column    may    be    taken    as 

hinged   at   that   point,   the   wind   acting 

upon    the    portion    below    the    point   of  I  ^       Hz, 

contra-flexure  being  neglected.    The  ver-  FIG.  83. 

tical   components   of  the   reactions  may 

now  be  found,  and,  since  the  horizontal  components  are  assumed 

to  be  equal,  the  reactions  themselves  may  be  determined. 

The  maximum  positive  bending  moment  in  the  column  is  at 
the  foot  of  the  knee-brace  of  the  leeward  column,  and  is  equal 


152  STRESSES    IN   A   TRANSVERSE   BENT.  Chap.  XV. 

to  H2  (d  —  y).  The  maximum  negative  moment  is  at  the  foot 
of  the  column,  and  is  equal  to  H2y  (see  Fig.  83). 

(3)  Columns  Fixed  at  Top  and  Base.  If  the  columns  are 
fixed  at  the  top  and  base,  the  point  of  contra-flexure  is  at  a  dis- 
tance y  — —  from  the  base  (see  Fig.  83).  In  this  case  the 

column  may  be  taken  as  hinged  at  the  point  of  contra-flexure, 
the  external  wind  below  this  point  being  neglected.  The  maxi- 
mum positive  moment  is  at  the  foot  of  the  knee-brace,  and  equals 

-j.  — —  •     and  the  maximum  negative  moment  is  at  the  base  of 

H2d 
the  column,  and  equals — . 

When  an  attempt  is  made  to  fix  the  columns,  the  resulting  con- 
dition probably  lays  between  that  shown  in  Case  2  and  that  shown 
in  Case  3.  It  is  seen  from  Case  2.  that  y  varies  but  slightly  with  a 
considerable  difference  in  the  ratio  of  d  to  h,  and  that  it  has  its 

minimum  value  of  -  when  h  — d.     In  Case  3  it  is  seen  that 
2 

y  =  -. .     The  assumption  commonly  made    when  some  effective 

means  are  used  to  fix  the  columns  at  the  base  is  that  y  =  — ,  and 

2 

this  assumption  will  be  made  in  this  text.  The  horizontal  com- 
ponents of  the  reactions  will  be  taken  equal,  and  the  vertical  com- 
ponents may  be  found  by  moments,  or  by  the  method  shown 

in  §  139- 

138.  Dead  and  Snow  Load  Stresses.  The  dead  and  snow 
load  stresses  in  the  truss  of  a  transverse  bent  are  the  same  as  for 
a  truss  supported  upon  masonry  walls.  If  the  columns  are  hinged 
at  the  top,  the  stresses  in  them  are  direct  compressive  stresses, 
caused  by  the  load?  upon  the  truss  and  by  the  weight  of  the  sides 
of  the  building  supported  by  the  columns.  If  the  columns  are 
fixed  at  the  top,  the  deflection  of  the  truss  will  produce  bending 
moments  in  the  columns  and  corresponding  stresses  in  the  knee- 
braces.  Since  the  deflection  of  the  truss  is  usually  quite  small, 


DEAD   AND    SNOW  LOAD    STRESSES. 


153 


the  bending  moments  in  the  columns  and  the  stresses  in  the  knee- 
braces  due  to  vertical  loads  will  be  neglected. 

The  dead  load  stress  diagram  for  a  transverse  bent  is  shown 
in  Fig.  84.  The  general  dimensions  of  the  building  and  of  the 
transverse  bent,  together  with  the  loads  used,  are  shown  in  the 
figure. 


Span  =  60-0".     Rise  =  I5'-0". 

Length  of  Building  =  90'-0". 

Distance  Between  Trusses  =  I5'-0". 

Height  of  Columns  =  20'-0"- 

Dead  Load  =  12,  Min-  Snow  Load  =10,  and 

Max-Snow  Load  =  20  Ibs-  per  sq-ft-  honproj- 


--*. 

b 


Dead  Load 
o       zooo    4000 


Max- Snow  Load 

0    2000  4000  6000 


rlinonow  Load 

0    1000  ZOOO  3000 


FIG.  84. 


Dead  and  Snow  Load 
Stress  Diagram 

STRESSES  IN  A  TRANSVERSE  BENT. 


The  snow  load  stresses  may  be  determined  by  proportion  from 
the  dead  load  stresses.  In  this  problem  the  maximum  snow,  which 
is  used  when  the  wind  load  is  not  considered,  is  taken  at  20  Ibs. 
per  sq.  ft.  of  horizontal  projection ;  and  the  minimum  snow  load, 
which  is  used  in  connection  with  the  wind  load,  is  taken  at  10  Ibs. 
per  sq.  ft.  of  horizontal  projection. 


154 


STRESSES    IN    A   TRANSVERSE   BENT. 


Chap.  XV. 


Since  the  deflection  of  the  truss  is  neglected,  there  are  no 
stresses  in  the  knee-braces  due  to  vertical  loads. 

139.  Graphic  Method  for  Determining  Wind  Load  Reac- 
tions.* The  following  is  a  very  convenient  graphic  method 
for  determining  the  wind  load  reactions  for  a  transverse  bent. 

Lay  off  2WN  (Fig.  85)  equal  to  the  total  normal  wind  load 
acting  upon  the  roof  area  supported  by  the  transverse  bent,  its 
point  of  application  being  at  the  center  of  the  length  RN.  Also, 
lay  off  2WH  equal  to  the  total  horizontal  wind  acting  upon  the 
side  area  supported  by  the  column  of  the  bent,  its  point  of  appli- 


FIG.  85.     GRAPHIC  METHOD  FOR  DETERMINING  REACTIONS. 

cation  being  at  the  center  of  the  column  length  AR.  Find  the 
resultant  LS  =  2W  of  the  normal  and  horizontal  wind  forces, 
and  produce  the  line  of  action  of  this  resultant  until  it  intersects 
at  B  a  vertical  line  through  N,  the  apex  of  the  truss.  Lay  off 
DB  =  BG  =  ^LS  (since  the  horizontal  components  of  the  reac- 
tions are  equal,  by  hypothesis).  Join  the  points  A  and  B,  also 
the  points  B  and  C,  and  from  the  points  D  and  G,  draw  the  ver- 

*This  graphic  method  of  determining  the  reactions  is  that  given  in 
Ketchum's  "Steel  Mill  Buildings." 


GRAPHIC    DETERMINATION    OF    REACTIONS.  155 

tical  lines  DE  and  GF,  respectively.  Then  ED  represents  the 
vertical  component  V2  of  the  right  reaction  R2,  and  GF  repre- 
sents the  vertical  component  V1  of  the  left  reaction  Rle 

This  construction  may  be  proved,  as  follows:  Taking  the 
moments  of  the  external  forces  about  the  point  C,  and  solving 
for  V,  we  have 


or,  since  BG  — (by  construction), 

4  (area  triangle  BGC)  (   . 

But  area  triangle  BGC  =  FG  X  L. 

4 

For,  area  triangle  BGC  =  area  triangle  BGF  +  area  triangle 
CGF 

FG  X  c       FG  X  d         FG  X  L 

-r  -+-—       —  (3) 

Substituting  this  value  of  the  area  of  the  triangle  BGC  in  equa- 
tion (2),  we  have 

V\  =  GF,  which  proves  the  construction. 

In  like  manner,  it  may  be  shown  that  ED  =  V3. 

The  left  reaction  Rx  =  GM  is  now  determined  by  finding  the 
resultant  of  Hj  and  V15  as  shown  in  the  figure.  The  right  reac- 
tion R2  =  MD  is  equal  to  the  resultant  of  H2  and  V2. 

In  the  case  shown,  the  columns  are  taken  as  hinged  at  the  base. 
If  the  columns  are  fixed  at  the  base,  the  above  construction 
should  be  modified  by  taking  the  columns  as  hinged  at  the  point 
of  contra-flexure  and  neglecting  the  wind  load  below  this  point. 

140.  Wind  Load  Stresses — Columns  Hinged  at  Base.  The 
wind  load  stress  diagram  for  a  transverse  bent  having  the  columns 
hinged  at  the  base  is  shown  in  Fig.  86.  The  dimensions  of  the 
bent  are  the  same  as  those  shown  in  Fig.  84.  The  wind  load  on 
the  roof  is  taken  at  22  Ibs.  per  sq.  ft.,  which  is  the  normal  com- 
ponent of  a  horizontal  wind  load  of  30  Ibs.  per  sq.  ft.  (see  Fig. 


156 


STRESSES    IN    A   TRANSVERSE   BENT. 


Chap.  XV. 


47,  Duchemin's  formula).  The  horizontal  wind  load  on  the  side 
of  the  building  is  taken  at  20  Ibs.  per  sq.  ft.  The  distribution  of 
the  loads  is  shown  on  the  truss  diagram,  and  the  reactions  are 
determined  by  the  method  shown  in  §  139.  Since  the  reactions  act 
at  the  bases  of  the  columns,  they  produce  a  bending  moment  at 


Nor-  Wind  ,  22  Ibs.  sq.ft. 
Hor.      „        20    „       „     „ 


R,^     IV, 
Columns   Hinged  at  Base  Max- Moment  =  H2d. 

W^-.jt--       — .      _       _      _ s~!\  ** 

.6 


Wind  Load 
Stress  Diagram 


5000 


10000. 


FIG.  86.     STRESSES  ix  A  TRANSVERSE  BENT — COLUMNS  HINGED. 

the  foot  of  the  knee-brace.  The  difficulty  of  drawing  the  stress 
diagram  due  to  this  moment  is  overcome  by  trussing  the  columns 
as  shown  in  the  figure.  The  stresses  obtained  for  the  members  of 
the  auxiliary  trusses  are  not  used,  and  the  resulting  stresses  in 
the  columns  are  not  true  stresses.  The  stresses  in  all  the  mem- 
bers of  the  truss  and  in  the  knee-braces  are,  however,  true  stresses. 


COLUMNS   HINGED   AT   BASE.  157 

The  complete  stress  diagram  is  shown  in  Fig.  86,  the  kind  of 
stress  in  each  member  being  indicated  by  arrows  in  the  truss 
diagram.  The  members  shown  by  dotted  lines  in  the  truss  dia- 
gram are  not  stressed  when  the  wind  acts  upon  the  left  side  of 
the  truss.  The  true  direct  stresses  in  the  windward  and  leeward 
columns  are  respectively  equal  to  V±  and  V2. 

The  maximum  bending  moment  occurs  at  the  foot  of  the 
knee-brace  of  the  leeward  column,  and  is  equal  to  H2d. 

The  unit  stress  in  the  extreme  fiber  of  the  column  due  to  the 
wind  moment  may  be  found  from  the  formula, 

_         My      .* 

-piJ' 

I  ±  

where 

M  =  maximum  bending  moment  in  inch-pounds ; 
y  =  distance   from  neutral  axis  to   extreme  fiber,  the  axis 
being  perpendicular  to  the  external  force  causing  the 
bending  moment; 

I  =  moment  of  inertia  of  the  section  of  the  number  about  an 
axis  perpendicular  to  the  direction  of  the  force  causing 
moment ; 

P  =  total  direct  loading  in  the  member  in  pounds ; 
L  =  length  of  member  in  inches. 

c  =  constant  depending  upon  condition  of  ends  of  member. 

For  a  member  hinged  at  both  ends,  use  c=io;    for 

member  fixed  at  one  end  and  hinged  at  the  other,  use 

c  =  24 ;  and  for  member  fixed  at  both  ends,  use  c  =  32 ; 

E  =  modulus  of  elasticity  of  the  member.     For  steel,  it  may 

be  taken  at  29000000. 

The  sign  is  to  be  plus  if  P  causes  compression,  and  minus  if 
P  causes  tension. 

141.  Wind  Load  Stresses — Columns  Fixed  at  Base.  The 
wind  load  stress  diagram  for  a  transverse  bent  with  the  columns 
fixed  at  the  base  is  shown  in  Fig.  87.  The  dimensions  of  the  bent 


•See "Theory  and  Practice  of  Modern  Framed  Structures,"  by  John- 
son, Bryan,  and  Turneaure. 


158 


STRESSES    IN    A   TRANSVERSE   BENT. 


Chap.  XV. 


are  the  same  as  shown  in  Fig.  84.    The  wind  load  on  the  roof  is 
taken  at  22  Ibs.  per  sq.  ft.,  and  on  the  sides,  at  20  Ibs.  per  sq.  ft. 

It  should  be  noted  that  in  this  case  the  columns  are  considered 
as  fixed  at  the  points  of  contra-flexure,  and  that  the  wind  below 


Nor  Wind,  ZZ  Ibs  sq-ft 
Hor-    »       20    »       •• 
Columns  Fixed  at  Base 


c1 


Wind   Load 
Stress  Diagram 


x    x 


5000 


10000 


FIG.  87.     STRESSES  IN  A  TRANSVERSE  BENT — COLUMNS  FIXED. 


these  points  is  neglected.     The  remaining  solution  is  similar  to 
that  shown  in  §  140. 

The  maximum  positive  bending  moment  is  at  the  foot  of  the 

knee-brace  of  the  leeward  column,  and  is  -f-.   ~  . .     The  maxi- 
/  2 

mum  negative  bending  moment  is   at  the   foot  of  the  leeward 

H.d 

column,  and  is  — i_. 

2 


COLUMNS    FIXED    AT    BASE.  159 

The  advantages  of  fixing  the  columns  are  shown  by  a  com- 
parison of  the  stress  diagrams  given  in  Fig.  86  and  in  Fig.  87 
and  of  the  maximum  bending  moments  for  the  two  cases.  Both 
stress  diagrams  are  drawn  to  the  same  scale. 

The  unit  stress  in  the  column  caused  by  the  bending  moment 
may  be  found  by  applying  the  formula  shown  in  §  140. 

Since  the  wind  may  act  from  either  side,  it  is  seen  that  many 
of  the  members  are  subjected  to  reversals  of  stress. 

The  maximum  and  minimum  stresses  caused  by  the  different 
loadings  may  be  found  by  making  the  combinations  indicated 
in  §  127. 


CHAPTER  XVI. 
MISCELLANEOUS  PROBLEMS. 

142.  Stresses  in  a  Grand  Stand  Truss.  It  is  required  to 
determine  the  stresses  in  the  grand  stand  truss  shown  in  Fig. 
88,  a.  In  this  structure,  the  knee-brace  and  the  seat  beam  meet 
at  the  point  M.  Since  the  left  column  is  braced  by  the  seat  beam, 
all  the  horizontal  component  of  the  wind  will  be  taken  by  this 
beam.  The  right  reaction  due  to  the  wind  loads  will  therefore  be 
vertical,  and  there  will  be  no  bending  moment  in  this  column  due 
to  the  wind.  The  wind  on  the  vertical  side  of  the  building  will 
not  be  considered,  as  it  will  be  resisted  by  the  seat  beam  and  will 
cause  no  stresses  in  the  truss.  The  general  dimensions  of  the 
truss,  together  with  the  loads  used,  are  shown  in  Fig.  88,  a. 

The  dead  load  stress  diagram  is  shown  in  Fig.  88,  b.  The 
reactions  R3  and  R3'  may  be  obtained  by  drawing  the  force  and 
funicular  polygons,  or  by  the  method  of  moments.  The  kind  of 
stress  is  indicated  by  the  arrows  in  the  stress  diagram,  compres- 
sion being  denoted  by  arrows  acting  away  from  each  other,  and 
tension  by  arrows  acting  toward  each  other.  In  this  problem,  the 
arrows  are  placed  on  the  stress  diagram,  instead  of  the  truss 
diagram,  to  avoid  confusion;  since  the  same  truss  diagram  is 
used  for  all  the  stress  diagrams. 

The  snow  load  stress,  if  considered,  may  be  determined  from 
'the  dead  load  stress  diagram. 

The  wind  load  stress  diagram,  considering  the  wind  as  acting 
towards  the  left,  is  shown  in  Fig.  88,  c.  The  reactions  R2  and  R2' 
are  determined  by  means  of  the  force  and  funicular  polygons. 
In  finding  these  reactions,  the  resultant  2WR  of  the  wind  load 'is 
used  instead  of  the  separate  loads  to  simplify  the  solution.  It  is 
seen  that  the  left  reaction  has  a  downward  vertical  component 

160 


STRESSES  IN  A  GRAND  STAND  TRUSS. 


161 


O1       5'      10'      15'     20' 


Dead  Load  =  10  Ibs-per  sq-ft-hor-proj 
Nor.  comp-of  wind  =  17  Ibs-per  sq-ft 
Distance  between  trusses  =  20ft- 


Wind  Load 
Stress  Diaqrafn 
Wind  Right 


Wind  Load 
Stress  Diaqram 

Wind   Left  (d)          8 

FIG.  88.     STRESSES  IN  A  GRAND  STAND  TRUSS. 


162 


MISCELLANEOUS    PROBLEMS. 


Chap.  XVI. 


due  to  the  wind  on  the  cantilever  side  of  the  truss.  The  stresses 
may  be  obtained  from  the  stress  diagram  shown  in  Fig.  88,  c. 

The  wind  load  stress  diagram,  considering  the  wind  as  acting 
towards  the  right,  is  shown  in  Fig.  88,  d.  The  reactions  Rl  and 
R/  are  determined  by  means  of  the  force  and  funicular  polygons 
as  shown,  the  resultant  of  the  wind  loads  on  the  left  being  used 
instead  of  the  separate  loads.  The  stresses  may  be  obtained  -from 
the  stress  diagram  shown  in  Fig.  88,  d. 

If  the  maximum  and  minimum  stresses  are  required,  they  may 


R 


H'=<p  H=^P 

---*-*• *»— «-2 H 

R      , »    P2   ,  ;x  P3   x   P4   x  PSJ 


(b) 


FIG.  89.  STRESSES  IN  A  TRESTLE  BENT. 

be  determined  by  combining  the  stresses  due  to  the  different 
loadings. 

143.  Stresses  in  a  Trestle  Bent.  It  is  required  to  find  the 
stresses  in  the  trestle  bent,  loaded  as  shown  in  Fig.  89,  a.  Since 
the  same  detail  is  used  at  the  bases  of  the  columns,  it  will  be 
assumed  that  the  horizontal  components  of  the  reactions  are 
equal. 

The  stress  diagram  for  the  bent,  shown  in  Fig.  89,  b,  is  drawn 


STRESSES    IN    A   TRESTLE   BENT. 


163 


by  starting  with  the  force  Px  acting  at  the  top  of  the  bent.  The 
diagram  is  completed  by  taking  each  of  the  horizontal  components 
of  the  reactions  at  the  base  equal  to  one-half  of  the  total  wind 
load  upon  the  structure.  The  stresses  may  be  obtained  from  the 
stress  diagram,  the  kind  of  stress  being  indicated  by  the  arrows. 


Moment  =  18000  x  3.4  =  61200  irvlbs. 
Direct  Shear  ,  5  *  18000  r  5  =  3600  Ibs 
"To  get  a  ,the  shear  due  to  moment 
at  a  units  distance  from  the 
center  of  gravity,  we  have 
Moment, M,=  afdf+df+df+di+dsJ 
or,  61  ZOO  =  a  (-9%  2-66f+ 1 842+  l-842t  Zbb*) 
Therefore,  a  =   £820. 

RESULTS 


Rivet 

d 

d2 

SM 

s 

R 

I 

.90 

.81 

2530 

3600 

6100 

z 

2.66 

7.07 

7500 

3600 

9300 

3 

1.84 

3.39 

5190 

3600 

3500 

4 

1-84 

3.39 

5190 

3600 

3500 

5 

2.66 

7.07 

7500 

3600 

9300 

SM  =  Shear  due  to  Moment  =  axd. 
5  =  Shear  due  to  direct  load,  P. 
R=   Resultant  Shear 

(a) 


(c) 

Force  Polygon 


FIG.  90.     ECCENTRIC  RIVETED  CONNECTION. 


THE 

UNJVERSITY 


164  MISCELLANEOUS   PROBLEMS.  x       Chap.  XVI. 

The  vertical  components  of  the  reactions  are  equal,  but  act  in 
opposite  directions. 

144.  Eccentric  Riveted  Connection.  It  is  required  to  find 
the  shearing  stress  in  each  of  the  five  rivets  in  the  connection 
shown  in  Fig.  90,  b.  The  spacing  of  the  rivets  is  that  used  in 
a  six-inch  angle  for  a  standard  channel  connection.  The  load  P 
is  transferred  to  its  connecting  member  at  the  left  edge  of  the 
angle.  Since  the  load  is  not  transferred  at  the  center  of  gravity 
of  the  connecting  rivets,  it  is  seen  that  there  is  a  tendency  for 
rotation,  which  causes  additional  shearing  stresses  in  the  rivets. 
The  total  shearing  stress  in  each  rivet  may  be  found  by  the 
following  method. 

Replace  the  force  P  by  an  equal  force  acting  through  the 
center  of  gravity  of  all  the  rivets  and  a  couple  whose  moment  M 
is  equal  to  P  multiplied  by  the  distance  from  its  line  of  action 
to  the  center  of  gravity  of  the  rivets.  In  this  example,  M  =  18  ooo 
X  3.4  =  61  200  in.  Ibs.  Now  the  force  P  acting  through  the  cen- 
ter of  gravity  will  cause  a  direct  shearing  stress  in  each  rivet 
equal  to  P  divided  by  the  number  of  rivets.  In  this  case,  the 
direct  shearing  stress  is  18  ooo  -4-5  =  3  600  Ibs.  The  moment  M 
of  the  couple  will  cause  a  shearing  stress  in  each  rivet,  which  may 
be  computed  as  follows :  Since  the  shear  in  each  rivet  due  to  the 
moment  will  vary  as  the  distance  of  that  rivet  from  the  center 
of  gravity  of  all  the  rivets,  it  follows  that  the  resisting  moment  of 
each  rivet  (which  is  equal  to  the  shear  in  the  rivet  multiplied  by 
*ts  moment  arm)  will  vary  as  the  square  of  its  distance  from  the 
center  of  gravity  of  all  the  rivets.  Now  if  a  represents  the  shear 
at  a  unit's  distance  from  the  center  of  gravity  due  to  the  moment 
M ;  and  dx,  d2,  etc.,  represent  the  distances  of  the  respective  rivets 
from  the  center  of  gravity,  the  following  relation  is  true : 
M  =  a  (d±2  +  d22  +  d32  +  d42  +  d52), 

M 

and  a~1S(:l2' 

Since  a  is  the  shear  at  a  unit's  distance  due  to  the  moment,  the 
shear  on  any  rivet  is  equal  to  a  multiplied  by  its  distance  from  the 
center  of  gravity  of  all  the  rivets.  The  total  shear  in  each  rivet 
may  now  be  determined  by  finding  the  resultant  of  the  shear  due 


ECCENTRIC   RIVETED   CONNECTION.  165 

to  the  direct  load  P  and  that  due  to  the  moment  of  the  couple. 
This  resultant  may  be  found  by  drawing  the  parallelogram  of 
forces,  As  shown  in  Fig.  90,  b. 

A  table  showing  the  important  data  for  this  problem,  together 
with  the  resultant  shear  on  each  rivet,  is  shown  in  Fig.  90,  a. 
This  table  shows  a  convenient  form  for  recording  results. 

The  force  polygon  for  the  shearing  forces  is  shown  in  Fig. 
90,  c,  and  the  funicular  polygon  in  Fig.  90,  d.  Since  both  poly- 
gons close,  the  forces  are  shown  to  be  in  equilibrium. 

Referring  to  the  resultant  shears  given  in  the  last  column  of 
the  table  shown  in  Fig.  90,  a,  it  is  seen  that  the  eccentric  connec- 
tion causes  large  shearing  stresses  in  some  of  the  rivets.  It 
therefore  follows  that  all  eccentric  connections  should  be  carefully 
investigated. 


PART  III. 

BEAMS. 

CHAPTER  XVII. 


BENDING  MOMENTS,  SHEAKS,  AND   DEFLECTIONS   IN  BEAMS 
FOR  FIXED  LOADS. 


This  chapter  will  treat  of  graphic  methods  for  determining 
bending  moments,  shears,  and  deflections  in  beams  for  fixed 
loads.  It  will  be  divided  into  three  articles,  as  follows:  Art.  i, 
Bending  Moments  and  Shears  in  Cantilever,  Simple,  and  Over- 
hanging Beams ;  Art.  2,  Graphic  Method  for  Determining  Deflec- 
tions in  Beams;  Art.  3,  Bending  Moments,  Shears,  and  Deflec- 
tions in  Restrained  Beams. 

Graphic  methods  may  be  readily  applied  to  the  determination 
of  bending  moments,  shears,  and  deflections,  and  in  many  cases 
afford  a  simpler  and  more  comprehensive  solution  than  algebraic 
processes,  especially  when  these  functions  are  required  at  several 
points  along  the  beam. 


ART.  i.     BENDING  MOMENTS  AND  SHEARS  IN  CANTILEVER,  SIM- 
PLE AND  OVERHANGING  BEAMS. 

145.  Definitions.  Vertical  forces,  only,  will  be  considered 
in  this  article ;  as  the  forces  acting  upon  a  beam  are  usually  ver- 
tical loads. 

Bending  Moment.  The  bending  moment  at  any  point,  or  at 

167 


168 


BENDING  MOMENTS  AND  SHEARS  IN  BEAMS. 


-  XVII. 


any  section,  of  a  beam  is  the  algebraic  summation  of  the  moments 
of  all  the  forces  on  one  side  of  the  point,  or  section.  The  bending 
moment  will  be  considered  positive  if  there  is  a  tendency  for  the 
beam  to  bend  convexly  downward,  and  will  be  considered  nega- 
tive of  there  is  a  tendency  for  the  beam  to  bend  convexly  upward. 

A  bending  moment  diagram  is  a  diagram  representing  the 
bending  moments  at  points  along  the  beam  due  to  the  given 
loading. 

Shear.  The  shear  at  any  section  of  a  beam  is  the  algebraic 
summation  of  all  the  vertical  forces  on  one  side  of  the  section. 
The  shear  is  positive  if  the  portion  of  the  beam  to  the  left  of  the 
section  tends  to  move  upward  with  reference  to  the  portion  to 
the  right  of  the  section.  The  shear  is  negative  if  the  portion  to 
the  left  tends  to  move  downward  with  reference  to  the  portion 
to  the  right  of  the  section. 

A  shear  diagram  is  a  diagram  representing  the  shears  at  points 
along  the  beam  due  to  the  given  loading. 

146.  Bending  Moment  and  Shear  Diagrams  for  a  Canti- 
lever Beam.  Two  conditions  of  loading  will  be  considered  for 
the  cantilever  beam,  viz.:  (a)  beam  loaded  with  concentrated 
loads,  and  (b)  beam  loaded  with  a  uniform  load. 

(a)      Cantilever    Beam    with    Concentrated    Loads.      It  is 


•I- 


Moment  Diaqram 
(b) 


Force  Polygon 
(a) 


D  D 

FIG.  91.     CANTILEVER  BEAM — :CONCENTRATED  LOADS. 


Art>   1-  CANTILEVER   BEAM.  169 

required  to  draw  the  bending  moment  and  shear  diagrams  for 
the  beam  MN  (Fig.  91),  loaded  as  shown  with  the  three  con- 
centrated loads  AB,  BC,  and  CD. 

Bending  Moment  Diagram.  To  construct  the  bending  moment 
diagram,  draw  the  force  polygon  (Fig.  91,  a),  assume  any 
pole  O,  and  draw  the  funicular  polygon  (Fig.  91,  b)  for  the 
given  loads.  The  diagram  shown  in  Fig.  91,  b  is  the  bending 
moment  diagram  for  the  beam  loaded  as  shown.  For  the  bend- 
ing moment  at  any  point  along  the  beam  is  equal  to  the  intercept 
under  that  point,  cut  off  by  the  funicular  polygon  and  the  hori- 
zontal line  m-n,  multiplied  by  the  pole  distance  H.  (See  §  64.) 

Referring  to  Fig.  91,  it  is  seen  that  the  maximum  bending 
moment  occurs  at  the  fixed  end  of  the  beam. 

Shear  Diagram.  To  construct  the  shear  diagram,  lay  off  the 
reaction  R  — AD  (Fig.  91,  c),  and  draw  the  horizontal  line  DD. 
Since  there  are  no  loads  between  the  left  reaction  and  the  load 
AB,  the  shear  is  constant  between  the  points  of  application  of 
these  forces,  and  is  represented  by  the  intercept  between  the  lines 
DD  and  AA.  At  the  point  of  application  of  the  load  BC,  the 
shear  is  reduced  by  the  amount  of  that  load.  The  shear  is  con- 
stant between  the  points  of  application  of  the  loads  AB  and  BC, 
and  is  represented  by  the  intercept  between  DD  and  BB.  Like- 
wise, the  shear  between  the  loads  BC  and  CD  is  constant,  and  is 
represented  by  the  intercept  between  DD  and  CC. 

Referring  to  the  cantilever  beam  shown  in  Fig.  91,  it  is  seen 
that  the  maximum  bending  moment  and  the  maximum  shear  both 
occur  at  the  same  point — the  fixed  end  of  the  beam. 

(b)  Cantilever  Beam  ivith  Uniform  Load.  It  is  required 
to  draw  the  bending  moment  and  shear  diagrams  for  the  beam 
MN  (Fig.  93),  loaded  as  shown  with  a  uniform  load. 

Bending  Moment  Diagram.  Referring  to  Fig.  92,  it  is  seen 
w-  Ibs-per  ft- 


A 


l-x 


PJ--MP 

FIG.  92. 


170  BENDING   MOMENTS   AND  SHEARS   IN   BEAMS.         Chap.  XVII. 

that  a  couple  is  required  to  fix  the  beam  at  the  left  end.     The 
magnitude  of  this  couple  may  be  found  as  follows : 

w  (1  — x)2 
Moment  of  forces  to  right  of  A=  -  — ,  (i) 

wx2 
and  moment  of  forces  to  left  of  A  =  —  Pa  +  Rx (2) 

which  are  the  equations  of  a  parabola. 

For  equilibrium,  the  sum  of  the  moments  of  the  forces  on  both 
sides  of  A  must  be  equal  to  zero,  and  the  magnitude  of  the  couple, 
which  is  Pa,  may  be  found  by  equating  (i)  and  (2)  to  zero  and 
solving  for  Pa,  noting  that  R  =  wl.  Then 

wx2         w  (1  —  x)2 
—  Pa  +  Rx —+—  —^-  —  =°> 

wl2 

or  Pa= ,  which  is  the  magnitude  of  the  couple  required  to 

2 

fix  the  beam  at  the  left  end.     It  is  seen  that  the  value  of  P 
depends  upon  the  arm  of  the  couple. 

wl2 
If  x  =  o,     M  =  —  Pa  = , 


and 


wl2 
if  x  =  1,     M  =  —  Pa  H =  o. 


To  construct  the  moment  diagram  for  the  beam  MN  (Fig.  93), 
divide  the  load  area  into  any  number  of  equal  parts  (in  this  case 
eight)  by  verticals,  and  take  the  weight  of  each  part  as  a  load 
acting  through  its  center  of  gravity.  Draw  the  force  polygon 
(Fig.  93,  a)  and  the  funicular  polygon  (Fig.  93,  b)  for  these 
loads;  and  trace  a  curve  (not  shown)  tangent  to  the  funicular 
polygon  at  its  ends  and  at  the  middle  points  of  its  sides.  The 
bending  moment  at  any  point  along  the  beam  is  equal  to  the  inter- 
cept, measured  between  the  horizontal  line  m-n  of  the  funicular 
polygon  and  the  curve,  multiplied  by  the  pole  distance  H.  The 
greater  the  number  of  parts  into  which  the  load  area  is  divided, 
the  more  nearly  will  the  funicular  polygon  approach  the  bending 
moment  parabola. 


Art.  1. 


CANTILEVER   BEAM. 


171 


Shear  Diagram.  The  shear  at  any  point  whose  distance  from 
the  left  end  of  the  cantilever  beam  is  x  may  be  represented  by 
the  equation,  S  =  R  —  wx.  If  x  =  o,  S  =  R ;  and  if  x  =  1, 
S  =  o.  Since  the  load  is  uniform,  the  shear  decreases  by  a  con- 
Uniform  load  of  w-  Ibs.  perlin-ft- 


Moment  Diagram 
(b) 


N1 


Shear  Diagram 

(O    * 


Force  Polygon 
(a) 


FIG.  93.     CANTILEVER  BEAM — UNIFORM   LOAD. 


stant  amount  towards  the  right  end  of  the  beam.  Therefore,  to 
draw  the  shear  diagram,  lay  off  AB  (Fig.  93,  c)  =  R,  and  draw 
the  horizontal  line  BC.  Join  the  points  A  and  C,  completing  the 
triangle  ABC,  which  is  the  required  shear  diagram. 

Referring  to  the  bending  moment  and  shear  diagrams,  it  is 
seen  that  neither  the  bending  moment  nor  the  shear  changes  sign 
throughout  the  length  of  the  beam. 

147.  Bending  Moment  and  Shear  Diagrams  for  a  Simple 
Beam.  The  bending  moment  and  shear  diagram  for  a  simple 
beam  will  be  constructed  for  two  conditions  of  loading,  viz. :  (a) 
beam  loaded  with  concentrated  loads,  and  (b)  beam  loaded  with 
a  uniform  load. 

(a)  Simple  Beam  with  Concentrated  Loads.  It  is  required 
to  draw  the  bending  moment  diagram  and  the  shear  diagram 


172 


BENDING   MOMENTS   AND   SHEARS   IN   BEAMS.         Chap.  XVII. 


for  the  beam  MN   (Fig.  94),  supported  at  its  ends,  and  loaded 
with  concentrated  loads  as  shown. 

Bending  Moment  Diagram.  To  construct  the  bending  moment 
diagram,  draw  the  force  polygon  (Fig.  94,  a),  assume  any  pole  O, 
and  draw  the  funicular  polygon  (Fig.  94,  b),  which  is  the  required 
bending  moment  diagram.  For,  the  bending  moment  at  any 


a  b      b 


M 


Id       die 

1  .1          N 


1  ,s=_ 


(a) 
Force  Polygon 


Shear  Diagravn 


FIG.  94.     SIMPLE  BEAM — CONCEXTKATED  LOADS. 

point  along  the  beam  is  equal  to  the  intercept  under  the  point  of 
moments  multiplied  by  the  pole  distance  H.  Referring  to  the 
bending  moment  diagram  (Fig.  94,  b),  it  is  seen  that  the  moment 
of  the  forces  to  the  left  of  any  point  along  a  simple  beam  is  posi- 
tive, and  does  not  change  its  sign  throughout  the  length  of  the 
beam.  In  this  particular  example,  it  is  seen  that  the  maximum 
bending  moment  occurs  at  the  point  of  application  of  the  load  BC. 
Shear  Diagram.  To  construct  the  shear  diagram,  lay  off 
FA  —  R!  (Fig.  94,  c),  and  draw  the  horizontal  line  FF.  Between 
the  left  reaction  and  the  load  AB,  the  shear  is  equal  to  Rx ;  there- 
fore from  A,  draw  the  horizontal  line  AA.  Then  the  intercept 
measured  between  FF  and  AA  represents  to  scale  the  shear 


Art.  1. 


SIMPLE    BEAM. 


173 


between  the  left  end  of  the  beam  and  the  load  AB.  Between  the 
loads  AB  and  BC,  the  shear  is  equal  to  Rx  —  AB ;  therefore  from 
A,  draw  the  vertical  line  AB,  representing  to  scale  the  force  AB, 
and  from  B,  draw  the  horizontal  line  BB.  Then  the  intercept 
between  BB  and  FF  represents  to  .scale  the  shear  between  the 
loads  AB  and  BC.  It  is  seen,  by  referring  to  Fig.  94,  c,  that  the 
shear  between  the  left  end  of  the  beam  and  the  load  BC  is  posi- 
tive. Between  the  loads  CD  and  DE,  the  shear  equals  R±  —  AB  — 
BC  —  CD,  and  is  represented  by  the  intercept  between  FF  and 
DD.  Between  the  load  DE  and  the  reaction  R2,  the  shear  equals 
Rt  —  AB  —  BC  —  CD  —  DE,  and  is  represented  by  the  intercept 
between  FF  and  EE.  The  shear  between  the  load  BC  and  the 
right  end  of  the  beam  is  negative. 

Referring  to  Fig.  94,  b  and  Fig.  94,  c,  it  is  seen  that  the  maxi- 
mum bending  moment  occurs  at  the  point  of  zero  shear,  i.  e.,  at 
the  load  BC. 

(b)  Simple  Beam  with  Uniform  Load.  It  is  required  to 
draw  the  bending  moment  and  shear  diagrams  for  the  beam  MN 
(Fig.  95),  supported  at  its  ends,  and  loaded  with  the  uniform 
load,  as  shown. 


Uniform  load  of  w  !bs-per  lin-  ft- 


M 


(c) 
Shear  Diagram 


Force  Polycjon 


FIG.  95.     SIMPLE  BEAM — ^UNIFORM  LOAD. 


174  BENDING   MOMENTS   AND   SHEARS   IN   BEAMS.         Chap.  XVII. 

Bending  Moment  Diagram.  To  construct  the  bending  moment 
diagram,  divide  the  load  area  into  any  number  of  equal  parts 
(in  this  case  eight)  by  verticals,  and  take  the  weight  of  each  part 
as  a  force  acting  through  its  center  of  gravity.  Draw  the  force 
polygon  (Fig.  95,  a)  for  these  loads,  assume  any  pole  O,  and 
draw  the  funicular  polygon  (Fig.  95,  b).  Trace  a  curve  (not 
shown  in  the  figure)  tangent  to  the  funicular  polygon  at  its  ends 
and  at  the  middle  points  of  its  sides.  Then  the  bending  moment 
at  any  point  along  the  beam  will  be  equal  to  the  intercept  under 
that  point,  between  the  curve  and  the  closing  line  of  the  funicular 
polygon,  multiplied  by  the  pole  distance  H. 

The  curve  drawn  tangent  to  the  middle  points  of  the  sides  of 
the  funicular  polygon  is  a  parabola,  and  the  greater  number  of 
parts  into  which  the  load  area  is  divided,  the  more  nearly  will  the 
funicular  polygon  approaching  the  bending  moment  parabola.  It 
will  now  be  shown  that  the  bending  moment  curve  for  a  uniform 
load  is  a  parabola. 

Let  L  =  span  of  beam,  w  =  weight  of  uniform  load  per  linear 
foot,  and  x  =  distance  from  the  left  support  to  the  point  of 
moments.  Then  the  bending  moment  at  any  point  whose  distance 

wx2      w 
from  the  left  end  is  x  is,  M  =  RjX =  —  ( Lx  —  x2 ) ,  which 

is  the  equation  of  a  parabola.    The  moment  is  a  maximum  when 
x  — — -,  and  is  equal  to  -JwL*. 

The  parabola  may  be  drawn  without  constructing  the  force 
and  funicular  polygons,  as  follows :  In  Fig.  95,  b,  lay  off  the 
ordinate  mn  =  nr  =  -JwL2  =  moment  at  the  center  of  the  beam ; 
and  connect  the  point  r  with  the  points  i  and  5.  Divide  the  lines 
ir  and  r5  into  the  same  number  of  equal  parts,  and  number  them 
as  shown  in  Fig.  95,  b.  Join  like  numbered  points  by  lines,  which 
will  be  tangents  to  the  required  parabola. 

Shear  Diagram.  To  construct  the  shear  diagram,  lay  off 
AC  =  R!  (Fig.  95,  c),  and  from  C,  draw  the  horizontal  line  CC. 
Also,  lay  off  CB  =  R2  downward  from  C,  and  connect  the  points 
A  and  B,  which  gives  the  required  shear  diagram. 

It  is  seen  from  Fig.  95,  b  and  Fig.  95  c  that  the  shear  is  zero 


Art.  1. 


OVERHANGING   BEAM. 


175 


at  the  center  of  the  beam,  and  that  the  moment  is  a  maximum  at 
the  point  of  zero  shear,  i.  e.,  at  the  center  of  the  beam. 

The  equation  expressing  the  shear  at  any  point  is  S  =  Rx  — 

wx  =  JwL  —  wx  =  w  ( x),  which  is  the  equation  of  the 

2 

inclined  line  AB  (Fig.  95,  c). 

It  will  now  be  shown  that  the  bending  moment  at  any  point 
along  a  simple  beam  is  the  definite  integral  of  the  shear  between 
the  point  in  question  and  either  point  of  support.  For, 


fx  /"x          /L  \  W,  2s 

I      S=l     w(  — -  —  x)—  — -  ( Lx —  x2  )  = 

Jo  J  o         \    2  /  2 


M. 


The  above  equation  shows  that  the  bending  moment  at  any 
point  in  a  simple  beam  uniformly  loaded  is  equal  to  the  area  of 
the  shear  diagram  on  either  side  of  the  point. 

148.  Bending  Moment  and  Shear  Diagrams  for  an  Over- 
hanging Beam.  Two  conditions  of  loading  will  be  considered 
for  the  overhanging  beam,  viz.:  (a)  beam  loaded  with  concen- 
trated loads,  and  (b)  beam  loaded  with  a  uniform  load. 

(a)      Overhanging   Beam   with    Concentrated   Loads.     It   is 


I  I  I  ^ 

ab  be         c  d          ,    die        elf        B 

M          J  +  M          P       +  k.  i  5 


(ej 
Shear  Diagram 


1_D 


PIG.  96.     OVERHANGING  BEAM — CONCENTRATED  LOADS. 


176  BEND1XG    MOMENTS   AND   SHEARS   IN    BEAMS.         Chap.  XVII. 

required  to  draw  the  bending  moment  and  shear  diagrams  for 
the  beam  MPN  (Fig.  96),  loaded  with  concentrated  loads. 

Bending  Moment  Diagram.  To  construct  the  bending  mo- 
ment diagram,  draw  the  force  polygon  (Fig.  96,  a),  assume  any 
pole  O,  and  draw  the  funicular  polygon  (Fig.  96,  b),  which  is 
the  required  bending  moment  diagram.  The  moment  at  any 
point  along  the  beam  is  equal  to  the  intercept  under  the  point 
multiplied  by  the  pole  distance  H. 

The  bending  moment  diagram  shown  in  Fig.  96,  b,  is  not  in 
a  convenient  form  for  comparing  moments  at  different  points, 
and  an  equivalent  diagram  will  now  be  drawn  which  shall  have 
all  intercepts  measured  from  a  horizontal  line.  If  the  line  mn 
(Fig.  96,  d),  common  to  the  two  outside  forces  R^  and  EF,  is 
made  horizontal,  it  is  seen  that  the  intercepts  will  all  be  meas- 
ured from  a  horizontal  line.  To  draw  the  bending  moment 
diagram .( Fig.  96,  d),  construct  the  new  force  polygon  (Fig.  96, 
c),  as  follows:  Since  mn  is  to  be  horizontal  and  common  to  the 
forces  Rj  and  EF,  draw  the  ray  O'F  (Fig.  96,  c)— H  (Fig. 
96,  a),  corresponding  to  the  string  mn ;  and  from  F  draw  Rt  = 
FA  (already  found)  upward.  From  A,  draw  in  succession  the 
loads  AB,  BC,  and  CD  downward  from  D;  and  from  D,  draw 
DD'=:R2  upward.  Then  from  D',  draw  the  loads  D'E  =  DE 
and  EF  downward,  closing  the  force  polygon  at  F.  Construct 
the  funicular  polygon  shown  in  Fig.  96,  d.  It  is  readily  seen, 
since  Rx  and  EF  have  a  common  point  F  in  the  force  polygon 
and  since  FO'  is  horizontal,  that  the  funicular  polygon,  or  bend- 
ing moment  diagram  (Fig.  96,  d),  will  have  intercepts  measured 
from  a  horizontal  line  equal  to  those  in  Fig.  96,  b. 

Referring  to  Fig.  96,  b  and  Fig.  96,  d,  it  is  seen  that  the 
bending  moment  for  this  problem  has  a  maximum  positive  value 
at  the  load  BC,  that  it  passes  through  zero  at  the  point  p,  and 
that  it  has  its  maximum  negative  value  at  P,  the  point  of  applica- 
tion Qf  the  reaction  R2. 

Shear  Diagram.  The  shear  diagram  for  the  overhanging 
beam  MN  loaded  with  concentrated  loads  is  shown  in  Fig.  96,  e. 
Referring  to  this  diagram,  it  is  seen  that  the  shear  is  positive  at 
the  left  end  of  the  beam,  that  it  passes  through  zero  at  the  load 


Art.   7. 


OVERHANGING   BEAM. 


177 


BC,  and  that  it  has  its  maximum  negative  value  between  the 
load  CD  and  the  reaction  R2.  It  is  further  seen  that  the  shear 
passes  through  zero  at  P,  the  point  of  application  of  the  reaction 
R2,  and  that  it  has  its  maximum  positive  value  between  the  reac- 
tion R2  and  the  load  DE. 

The  maximum  moment  occurs  at  the  point  of  zero  shear,  and 
since  the  shear  passes  through  zero  at  two  points,  the  moment 
will  have  maximum  values  at  these  points — -one  of  which  is  the 
maximum  positive  moment,  and  the  other,  the  maximum  negative 
moment. 

Uniform  load  of  w  ibs-per  I'm -ft- 


Moment  Diaqrams 

!«*.; 


(e) 
Shear  Diagram 


FIG.  97.     OVERHANGING  BEAM — UNIFORM  LOAD. 

(b)  Overhanging  Beam  with  Uniform  Load.  It  is  required 
to  draw  the  bending  moment  and  shear  diagrams  for  the  beam 
MN  (Fig.  97).  The  beam  overhangs  both  supports,  and  is  loaded 
with  a  uniform  load. 

Bending  Moment  Diagram.  To  draw  the  bending  moment 
diagram,  divide  the  load  area  into  any  number  of  equal  parts  (in 
this  case  nine),  and  assume  the  weight  of  each  part  as  a  load 
acting  through  its  center  of  gravity.  Draw  the  force  polygon 
(Fig.  97,  a),  and  the  funicular  polygon  (Fig.  97,  b).  Trace  a 
curve  (not  shown  in  the  figure)  tangent  to  this  funicular  polygon 


178  DEFLECTIONS   IN   BEAMS.  Chap.  XV II. 

at  its  ends  and  at  the  middle  points  of  its  sides,  which  will  be  the 
required  bending  moment  diagram. 

The  equivalent  funicular  polygon  showing  the  intercepts  meas- 
ured from  a  horizontal  line  is  shown  in  Fig.  97,  d,  and  the  force 
polygon  is  shown  in  Fig.  97,  c.  If  a  curve  is  drawn  tangent 
to  this  funicular  polygon  at  its  ends  and  at  the  middle  points  of 
its  sides,  the  diagram  will  be  the  required  bending  moment  dia- 
gram for  the  given  loads  and  beam.  The  bending  moment  at 
any  point  is  equal  to  the  intercept  under  the  point  of  moments 
multiplied  by  the  pole  distance  H  =  H'. 

Referring  to  Fig.  97,  d,  it  is  seen  that  the  bending  moment  is 
negative  at  both  supports,  and  that  it  passes  through  zero  between 
the  supports  and  becomes  positive. 

Shear  Diagram.  The  shear  diagram  for  the  overhanging 
beam  loaded  with  a  uniform  load  is  shown  in  Fig.  97,  e.  It  is 
seen  that  the  shear  between  the  left  end  of  the  beam  and  the  left 
support  is  negative,  and  that  it  passes  through  zero  at  the  left 
support  and  becomes  positive.  The  shear  passes  through  zero  at 
p  and  becomes  negative.  At  the  right  support  it  again  passes 
through  zero  and  becomes  positive,  and  is  positive  between  the 
right  support  and  the  right  end  of  the  beam. 

The  shear  passes  through  zero  at  three  points,  therefore  the 
bending  moment  has  maxima  at  these  points.  The  maximum 
negative  moments  are  at  the  supports,  and  the  maximum  positive 
moment  is  between  the  supports. 


ART.  2.     GRAPHIC  METHOD  FOR  DETERMINING  DEFLECTIONS  IN 

BEAMS. 

149.  Explanation  of  Graphic  Method  — Constant  Moment 
of  Inertia.  The  deflection  at  any  point  along  a  beam  may  be 
readily  determined  graphically,  and  a  graphic  method  for  finding 
the  deflections  will  now  be  explained.  The  graphic  method  is 
especially  useful  when  the  deflections  are  required  at  several 
points  along  the  beam. 

Let  MN  (Fig.  98,  a)  be  a  horizontal  beam,  supported  at  its 


GRAPHIC    DETERMINATION    OF    DEFLECTIONS. 


179 


ends,  and  let  the  beam  be  divided  into  a  sufficient  number  of  parts 
(in  this  case  four)  that  the  polygon  representing  its  neutral  sur- 


FIG.  98.     GRAPHIC  DEFLECTIONS. 

face  may  very  closely  approximate  the  elastic  curve.     By  the 

elastic  curve  is  meant  the  curve  assumed  by  the  neutral  surface 

of  the  beam  when  the  elastic  limit  of  the  material  is  not  exceeded. 

Let  ab   (Fig.  98,  c)   be  the  position  taken  after  flex'ure  by  the 

neutral  surface  of  the  segment  at  the 

left  end  of  the  beam.    Also  let  Fig.  99 

represent  a  portion  of  a  beam  which 

has  deflected  as  shown.     The  angle  a 

(Fig.  98,  c)  between  be  and  b^  (the 

projection  of  ab)  represents  the  angle 

of  rotation  of  the  section   CD    (Fig. 

99)  to  C'D',  originally  parallel  to  AB. 

The  angle  a  (Fig.  98,  c)  also  equals 

the    angle    a    between    mn    and    mo 

(Fig.  99). 

Let  ab  =  dl  (Fig.  98,  c) .  Then  the  angle  a  may  be  found  from 

Mdl 
the  formula  tan  a  =  -™^»  where 


M  =  bending  moment  at  b,  in  inch-pounds, 
dl  — ab  or  b, 


180         .  DEFLECTIONS    IN    BEAMS.  Chap.  XVII. 

E  =  modulus  of  elasticity  of  the  material,  in  pounds, 
I  =  moment  of   inertia  of  the   cross   section  of  the  beam, 
which  in  this  case  is  assumed  to  be  constant  throughout  its  length. 
This  formula  may  be  deduced  as   follows :     The  stresses  at 
any  point  in  the  beam  shown  in  Fig.  99  will  vary  as  the  distance 
of  the  point  from  the  neutral  axis.     From  the  similar  triangles 
Omn  and  DnD'  (Fig.  99),  we  have 

R  :c  ::dl  :  A, 

or,  RA=cdl.  (i) 

Now  let  S  =  stress  on  extreme  fiber,  and  let  E  =  modulus  of 
elasticity  of  the  material.    Then 

A  :  S  : :  dl  :  E, 

Sdl 
or,  :— .  (2) 

Substituting  this  value  of  A  in  equation  (i),  and  solving  for  R, 
we  have 

EC 

R=nr.  (3) 

But  from  the  common  theory  of  flexure,  we  have 

Me 
— .  (4) 

Substituting  this  value  of  S  in  equation  (3),  we  have 

El  ,   N 

R==-rr-  (5) 

M 

dl 
From  Fig.  99,  it  is  seen  that  tan  a  ==  —5".  (6) 

Substituting  the  value  of  R  found  In  equation  (5)  in  equation  (6), 
we  have 

Mdl 
tana=    -jr-p  (7) 

Calculate  the  value  of  the  angle  a  from  equation  (7),  and 
draw  be.  The  angles  ax  and  a2  may  be  found  in  like  manner.  Cal- 
culate the  values  of  these  angles,  lay  them  off  at  c  and  d  (Fig. 
98),  respectively,  and  draw  the  lines  cd  and  de.  Draw  the  closing 
line  ae,  and  the  vertical  ordinates  will  have  their  true  values  in 


^.  #•  DEFLECTIONS   IN   A   SIMPLE  BEAM.  181 

either  Fig.  98,  b  or  Fig.  98,  c.  For  in  both  Fig.  98,  b  and  Fig. 
98,  c,  we  have 

at  B,     Bb  =  Bb, 
at  C,     Cc  =  CCi-cq  =  2  Bb-cq, 
at  D,     Dd  =  Dda-didjj-ddi  =  3  Bb-2  cc^ddi, 
and  at  e,    o  =  ee3-e2e8-e1e2-ee1  =  4  Bb~3  cq-2  dd^ee^ 
Since  cca,  dd1?  and  ee^  are  equal  in  both  Fig.  98,  b  and  Fig. 
98,  c ;  therefore  Bb,  Cc,  and  Dd  must  also  be  equal  in  both  figures. 
It  is  thus  seen  that  the  closing  line  ae  may  be  horizontal  or 
inclined  without  changing  the  values  of  the  deflection  intercepts. 
Referring  to  Fig.  98,  it  is  seen  that  the  successive  triangles 
aBb,  bccj,  cdd±,  and  dee!  have  one  side  equal  in  each  successive 
pair  of  triangles.     These  triangles  may  therefore  be  placed  in 
contact,  as  shown  in  Fig.  98,  d  and  Fig.  98,  e,  Fig.  98,  d  cor- 
responding to  Fig.  98,  b,  and  Fig.  98,  e  to  Fig.  98,  c.     The  tri- 
angles shown  in  Fig.  98,  d  and  Fig.  98,  e  suggest  another  method 
of  drawing  the  polygons  representing  the  neutral  surface  of  the 
beam.    Thus  with  a  pole  distance  dl  equal  to  one  division  of  the 
span,  lay  off  in  succession  the  distances  ccx,  ddlf  and  ee^  com- 

Mdl 

puted  from  the  formula  dl  tan  a  =  dl  -==-.     Draw  the  rays,  and 

construct  the  funicular  polygons  shown  in  Fig.  98,  b  and  Fig. 
98,  c. 

150.  Practical  Application.  The  method  explained  in 
§  149  will  now  be  applied  to  a  practical  problem. 

Let  MN  (Fig.  100 )  represent  a  12  in.  X  31-5  tt>.  X  40  ft.  I 
beam,  supported  at  its  ends,  and  sustaining  a  load  of  4000  pounds 
at  a  point  15  feet  from  the  left  end  of  the  beam.  It  is  required  to 
draw  the  elastic  curve  representing  its  neutral  surface  and  find  the 
magnitude  of  its  maximum  deflection. 

Draw  the  force  polygon  shown  in  Fig.  100,  a  and  the  funicular 
polygon  shown  in  Fig.  100,  "b.  Divide  the  span  into  any  number 
of  equal  parts  (in  this  case  eight)  by  verticals  in  the  moment 
diagram,  and  bisect  each  of  these  laminae  by  a  vertical,  drawn 
between  the  closing  line  cd  and  the  funicular  polygon  ced.  Con- 
sider these  verticals  as  loads,  and  lay  them  off  successively  on  the 


182 


DEFLECTIONS   IN    BEAMS. 


Chap.  XVII. 


vertical  load  line  XY  (Fig.  100,  c).  Assume  any  pole  O',  -draw 
rays,  and  construct  a  new  funicular  polygon  fgh  (Fig.  100,  d) 
with  the  closing  line  fg.  Draw  a  curve  tangent  to  the  middle 


h          (dj 
Deflection  Diagram 


FIG.  100.     DEFLECTIONS  IN  A  SIMPLE  BEAM — CONCENTRATED  LOAD. 

points  of  the  sides  of  this  funicular  polygon.  This  curve  will 
then  represent  the  elastic  curve  of  the  beam  in  an  exaggerated 
form,  and  the  intercepts  between  the  funicular  polygon  and  its 
closing  line,  which  represent  the  deflections  of  the  beam,  are 
each  to  be  divided  by  a  constant  to  obtain  their  true  values. 

The  value  of  this  constant  will  now  be  determined.    Referring 
to  Fig.  98,  b  and  Fig.  98,  c,  it  is  seen  that  the  ordinates  ccj,  ddt, 

Mdx2 
and  eel  are  each  equal  to  dx  tan  a,  which  may  be  written    -^ — ; 

since  tan  a  has  been  shown  equal  to  . ,  and  dl  may  be  taken 

equal  to  dx  for  the  elastic  curve  with  a  pole  distance  dx.  Now  in 
Fig.  101,  a,  Jay  off  successive  distances  on  the  deflection  load  line 

M 

equal  to  YF>  where  H  =  first  pole  distance ;    and  then  construct 
xl 

Fig.  101,  b.  It  is  now  seen  that  Fig.  101,  a  corresponds  to  Fig. 
loo,  c,  and  Fig.  101,  b  to  Fig.  100,  d.  From  the  similar  triangles 


Art. 


DEFLECTIONS   IN   A   SIMPLE   BEAM. 


183 


bcc,  and  O'i2,  cq :  ~  : :  dx  :  H',  from  which  CCl  =  ^dx     A  gim_ 
•H-  HH' 

ilar  relation  is  true  of  the  intercepts  ddt  and  ee^    Comparing  the 
values  obtained  for  the  verticals  cq,  ddx,  and  ee:  in  Fig.  101,  a 


with  the  true  verticals  cq,  dd1?  and  eex  in  Fig.  98,  b  and  Fig. 
98,  c,  we  find  the  following  relation, 

Mdx2       Mdx        HH'dx 


EI      km/      EI 

Therefore  the  true  verticals  may  be  found  from  those  in  Fig.  101 
by  multiplying  the  intercepts  measured  in  this  figure  by 

HH'dx 
EI 


HH'dx  u          Mdx2       Mdx 
~'    °r'~== 


The  sarrte  relation  is  true  for  the  actual  deflection  intercepts 
Bb,  Cc,  and  Dd,  which  correspond  to  those  in  Fig.  100,  d ;  since 
these  are  each  composed  of  the  elements  cclf  ddt,  and  eet,  as  has 
been  previously  shown. 

In  the  problem  given,  H  =  7  ooo  pounds,  H'  =  20  feet  =  240 

480 
inches,  dx  =    '    =  60  inches,  the  maximum  intercept  =  6.8  feet 

o 

=  81.6  inches,   1  =  215.8,   and   £  =  29000000.     Therefore  the 

,    a      *.'  A      /•       •      t,      \          8l.6X7000X240X6o 

maximum  deflection   A    (in  inches)  = 1 

215.8  X  29000000 

=  1.31. 

If  the  constant  by  which  each  measured  intercept  is  to  be  mul- 


184  DEFLECTIONS    IN    BEAMS.  Chap.  XVII. 

T-f  "FT'dx 
tiplied  is  used  in  the  form ,   then  the  measured  intercept, 

H',  and  dx  must  all  be  expressed  in  inches. 

If  the  measured  intercept,  H',  and  dx  are  expressed  in  feet, 
then  the  constant  should  be 

1728  HH'dx 

~EI~ 

The  method  explained  above  may  be  used  to  find  the  deflec- 
tion at  any  point  of  a  simple  beam,  a  cantilever  beam,  or  a  beam 
overhanging  the  supports,  sustaining  either  concentrated  or  uni- 
form loads.  If  the  length  of  the  parts  into  which  the  beam  is 
divided  is  small,  this  method  gives  results  sufficiently  accurate 
for  practical  purposes. 

151.  Deflection  Diagram  —  Variable  Moment  of  Inertia. 
The  method  of  determining  deflections,  described  in  §  149  and  § 
150,  is  applicable  to  beams  having  a  constant  moment  of  inertia. 
For  a  simple  beam,  the  bending  moment  increases  towards  the 
center,  and  for  an  economical  design  of  built-up  beams,  it  is 
necessary  to  increase  the  moment  of  inertia  of  the  beam  cor- 
respondingly. When  the  section  of  the  beam  is  not  constant,  the 
method  of  determining  deflections  should  be  modified;  as  will 
now  be  shown  by  the  solution  of  a  practical  problem.  The  solu- 
tion which  will  now  be  shown  gives  accurate  results. 

It  is  required  to  draw  the  deflection  diagram  for  the  plate 
girder  shown  in  Fig.  102.  The  span  of  the  girder  is  62  feet  4 
inches,  center  to  center;  the  distance,  back  to  back  of  angles,  is 
6  feet  2  inches ;  and  the  girder  is  loaded  with  a  uniform  load  of 
3  600  pounds  per  linear  foot.  The  section  of  the  beam  at  the 
different  points  is  as  shown  in  the  lower  part  of  Fig.  102.  The 
moment  of  inertia  is  increased  towards  the  center  of  the  girder 
by  the  addition  of  cover  plates. 

Draw  the  force  polygon  (Fig.  102,  a)  for  the  given  uniform 
load.  In  this  case,  one-half  the  total  uniform  load  is  assumed 
to  be  divided  into  five  parts,  and  these  partial  loads  are  assumed 
to  act  through  the  center  of  gravity  of  each  part.  Assume  any 
pole  O  and  pole  distance  H,  and  draw  the  funicular  polygon 


Art. 


DEFLECTIONS  IN  A  PLATE  GIRDER. 


(Fig.  102,  b).  A  curve  tangent  to  this  polygon  will  give  the 
moment  diagram.  Since  the  girder  is  symmetrical  about  its  cen- 
ter line,  the  moment  and  deflection  diagrams  need  be  drawn  for 
only  one-half  of  the  span. 

Uniform    load  of  3600  Ibs-  per  lin-  ft-  of  span. i 


f  I  Cover  P|.  14"xax49-0" 
FLANGE  1  I  Cover  PI-  14'xi"*  59-0" 
SECTION  ]  I  Cover  PI-  I4"xi"x  28-0" 

I  2  l»  -  6"x6"xi"x64-0" 
WEB  PL-  -  73i"x'  -  -  -  -  - 


14=15=  120560 
Is  =  100080 
Ii=  80130 
Ii  =  60700 


FIG.  102.     DEFLECTIONS   IN  A  PLATE  GIRDER. 


0.37" 


To  draw  the  deflection  diagram,  instead  of  dividing  the 
moment  area  by  equidistant  verticals  as  was  done  in  §  150,  divide 
it  into  laminae  by  verticals  dropped  from  the  ends  of  the  cover 


186  DEFLECTIONS   IN   BEAMS.  Chap.  XVII. 

plates.  The  moments  of  inertia  between  these  verticals  will  then 
be  constant.  Compute  the  area  of  each  lamina  in  square  feet 
(using  scale  of  beam),  locate  its  center  of  gravity,  and  assume 
a  force  numerically  equal  to  the  area  of  the  lamina  to  act  through 
its  center  of  gravity.  Compute  the  moments  of  inertia  of  the 
different  sections  of  the  beam.  The  values  of  these  moments  of 
inertia  for  the  .given  girder  are  shown  in  the  lower  part  of  Fig. 
102.  Determine  the  ratio  of  the  moment  of  inertia  of  each  sec- 
tion to  that  at  the  end  of  the  beam.  These  ratios  (reading  from 
the  end  of  the  beam)  are  equal  to  i.oo,  1.32,  1.65,  and  1.99, 
respectively.  Lay  off  the  moment  area  to  any  convenient  scale 
on  the  load  line  CD  (Fig.  102,  c).  From  D,  draw  a  horizontal 
line  DO4,  and  on  this  horizontal  line,  lay  off  the  pole  distance  Ht, 
using  the  same  scale  as  for  the  load  line  CD.  This  pole  distance 
may  be  taken  of  any  convenient  length.  Now  the  deflection  at 
any  point  varies  inversely  as  the  moment  of  inertia  of  the  section 
at  that  point.  Therefore  to  draw  a  deflection  diagram  whose 
intercepts  shall  bear  a  constant  ratio  to  the  true  deflections,  it  is 
necessary  to  increase  the  pole  distances  in  the  same  ratio  that  the 
moments  of  inertia  are  increased.  Lay  off  H2  (Fig.  102,  c)  = 

—  XH15     H3=1^XH1;     and  H4  =  H5=-f  XH,. 

-1!  Al  Al 

Join  the  point  4  with  the  point  O4.  Also,  since  the  moment 
of  inertia  is  constant  for  the  two  center  moment  areas,  join  the 
point  3  with  the  same  point  O4.  From  the  point  7,  draw  7~O3 
vertical,  to  intersect  3~O4  at  O3 ;  and  connect  the  points  2  and  O3. 
From  the  point  6,  draw  6-O2  vertical,  to  intersect  2-O3  at  O2 ; 
and  join  the  points  I  and  O2.  Also,  from  the  point  5,  draw  5-Oj 
vertical,  to  intersect  i-O2  at  O, ;  and  join  the  points  C  and  O^ 
Using  the  lines  C-O^  i-O^  2-O2,  3~O3,  4~O4,  and  D-O5  as  rays, 
draw  the  funicular  polygon  (Fig.  102,  d).  Trace  a  curve  tangent 
to  the  polygon,  which  will  give  the  required  deflection  diagram. 
To  get  the  maximum  deflection  A  (at  the  center),  multiply 

1728  X  H  X  HI  X  dx 

the  measured  intercept  y  by  — -  .    In  this  case 

El 

dx  =  i ;  since  the  area  was  taken  instead  of  the  intercept  in  the 


Art-  3-  RESTRAINED   BEAMS.  187 

moment  diagram.  Hl  is  always  to  be  taken  as  the  least  value  for  the 
moment  of  inertia.     Therefore 

1728  X  240000  X  150  X  10.6 

29000000X60700  =0-37  inch. 

The  deflection  Ax  at  any  other  point  along  the  girder,  where 
the  intercept  is  ylt  may  be  found  by  proportion,  i.  e.,  if  yx  repre- 
sents the  intercept  at  the  center  of  the  beam,  then 
Y!  :y  ::  At  10.37, 

A         0.37  X  y, 
or  AI= -.     This  requires  less  work 

than  a  second  substitution  in  the  formula. 


ART.    3.     BENDING   MOMENTS,    SHEARS,   AND   DEFLECTIONS    IN 
RESTRAINED  BEAMS. 

The  following  examples  of  restrained  beams  will  be  taken  up 
in  this  article,  viz. :  ( i )  cantilever  beam — a  beam  fixed  at  one 
end  and  free  at  the  other;  (2)  beam  fixed  at  one  end  and  sup- 
ported at  the  other;  (3)  beam  fixed  at  both  ends. 

152.  Definitions.     A  restrained  beam  is  a  beam  fastened 
at  one  or  more  points  in  such  a  manner  that  the  beam  is  not  free 
to  deflect  at  these  points. 

A  beam  is  fixed  at  any  point  if  its  neutral  surface  at  that  point 
is  horizontal. 

The  bending  moment  and  shear  diagrams  for  a  cantilever 
beam  may  be  drawn  without  first  finding  the  moment  of  the 
fixing  couple.  In  other  restrained  beams,  it  is  necessary  to  first 
find  the  deflections  at  the  ends,  considering  the  beam  as  sup- 
ported at  both  ends,  and  then  to  determine  the  value  of  the  fixing 
moment  that  will  make  the  neutral  surface  horizontal  at  the  fixed 
points. 

Referring  to  the  overhanging  beam  shown  in  Fig."  97,  it  is 
seen  that  the  overhanging  portions  of  the  beam  may  be  made 
of  such  lengths  that  the  beam  will  be  fixed  at  its  supports. 

153.  (i)     Bending  Moment,   Shear,   and   Deflection   Dia- 


188 


MOMENTS,   SHEARS,   AND   DEFLECTIONS.         Chap.  XVII. 


grams  for  a  Cantilever  Beam.  It  is  required  to  draw  the  bend- 
ing moment,  shear,  and  deflection  diagrams  for  the  cantilever 
beam  MN  (Fig.  103). 


Moment  Diagram 
(b) 


Shear  Diagram 
1     (c) 


Deflection    Diagram 
(e) 


FIG.  103.     CANTILEVER  BEAM — CONCENTRATED  LOADS. 

The  bending  moment  and  shear  diagrams  are  constructed  by 
the  methods  explained  in  Art.  I.  The  bending  moment  diagram 
is  shown  in  Fig.  103,  b,  and  the  shear  diagram  in  Fig.  103,  c. 

To  draw  the  deflection  diagram,  use  the  method  explained  in 
Art.  21.  Divide  the  moment  area  (Fig.  103,  b)  into  segments  by 
verticals,  and  lay  off  the  lengths  of  these  intercepts  on  a  vertical 
load  line  (Fig.  103,  d).  Take  any  pole  O',  and  construct  the 
funicular  polygon  shown  in  Fig.  103,  e.  Trace  a  curve  tangent 
to  this  funicular  polygon,  which  is  the  required  deflection  dia- 
gram. 

To  get  the  actual  deflection  at  any  point,  multiply  the  intercept 
between  the  curve  and  the  horizontal  line  by  the  constant  deter- 
mined by  equation  (i),  or  equation  (2),  §  150. 

154.  (2)  Bending  Moment,  Shear,  and  Deflection  Dia- 
grams for  Beam  Fixed  at  One  End  and  Supported  at  the 
Other.  The  diagrams  will  be  drawn  for  the  beam  loaded  (a) 


Art.  3. 


CANTILEVER    BEAM. 


189 


with  concentrated  loads,  and  (b)  with  a  uniform  load.  The  con- 
struction will  first  be  taken  up  in  detail,  and  the  proof  given ; 
after  which  simplified  constructions  will  be  shown. 


M 


a  b 


be 


f  . 


F1 


Shear  Diagram 
fc) 


Deflection  Diagram  -  Simple  i>am 
(e) 


Moment  Diagram- Fixing  Couple 
(f)  f. 


Deflection  Diagram -Fix  ing  Couple 
(h) 


Deflection  Diagram -Fixed  and  Supported 

(i) 

FIG.  1-04.     BEAM  FIXED  AND  SUPPORTED — CONCENTRATED  LOADS. 

(a)    Beam  Fired  and  Supported — Concentrated  Loads.     It 
is  required  to  draw  the  bending  moment,  shear,  and  deflection 


190  MOMENTS,   SHEARS,   AND   DEFLECTIONS.         Chap.  XVII. 

diagrams  for  the  beam  MN  (Fig.  104).  The  beam  is  fixed  at 
the  right  end  and  is  supported  at  the  left,  and  is  loaded  with  the 
three  concentrated  loads,  as  shown. 

Assume  that  the  beam  is  supported  at  both  ends,  and  draw 
its  bending  moment,  shear,  and  deflection  diagrams,  as  in  §  150. 
Also  draw  the  dividing  ray  O'm  (Fig.  104,  d)  parallel  to  the 
closing  string  eh  (Fig.  104,  e).  The  force  polygon  is  shown  in 
Fig.  104,  a;  the  bending  moment  diagram  in  Fig.  104,  b;  the 
shear  diagram  in  Fig.  104,  c ;  the  deflection  force  polygon  in  Fig. 
104,  d ;  and  the  deflection  diagram  in  Fig.  104,  e. 

Since  the  beam  is  to  be  fixed  at  the  right  end,  assume  that  a 
couple  whose  moment  is  sufficient  to  make  the  neutral  surface 
horizontal  is  applied  at  the  right  end  of  the  beam.  Let  the  tri- 
angle F'G'K'  (Fig.  104,  f)  be  the  moment  area  of  the  couple 
required  to  fix  the  beam  at  this  point.  The  moment  of  this  fixing 
couple  is  equal  to  wH ;  where  H  is  the  first  pole  distance,  and  w 
is  the  intercept  under  the  fixed  end  of  the  beam,  this  intercept 
being  the  altitude  of  the  moment  area  for  the  fixing  couple.  The 

T^ol' 
value  of  w  may  be  computed  from  the  formula  w  =      '"'    ;   where 

S  =  span  of  beam  in  feet,  1'  =  length  of  one  division  of  the 
moment  area  in  feet,  and  v2  =  length  m~7  (Fig.  104,  d).  It 
is  seen  that  w  may  be  found  from  the  above  formula  as  soon  as 
the  deflection  diagram  for  the  beam  treated  as  supported  at  both 
ends  has  been  drawn.  Compute  w,  and  construct  the  bending 
moment  triangle  F'G'K'  (Fig.  104,  f).  Then  wH  is  the  moment 
of  the  fixing  couple  at  N,  and  the  moment  at  any  point  along  the 
beam  due  to  the  fixing  couple  is  equal  to  the  intercept  under  that 
point  multiplied  by  H. 

Construct  the  deflection  diagram  for  the  fixing  couple  shown 
in  Fig.  104,  h.  The  force  polygon,  containing  the  intercepts  in 
the  moment  triangle  (Fig.  104,  f),  is  shown  in  Fig.  104,  g,  the 
pole  distance  H2  being  taken  equal  to  H' ;  and  the  deflection 
diagram  is  shown  in  Fig.  104,  h.  Since  the  moment  area  shown 
in  Fig.  104,  f  is  that  for  the  fixing  couple,  it  is  seen  that  fg  (Fig. 
104,  e)  must  be  equal  to  fg'  (Fig.  104,  h),  i.e.,  the  deflection 


•Art.  3.  BEAM   FIXED  AND  SUPPORTED.  191 

caused  by  the  fixing  couple  must  be  equal  and  opposite  to  that 
at  the  end  of  the  beam  supported  at  both  ends. 

It  will  now  be  shown  that  the  altitude  G'K'   (Fig.   104,  f) 

3v.,l' 
=  w  =  — F— •    Assume  that  the  verticals  in  the  triangle  F'G'K' 

o 

were  drawn  one  foot  apart.  The  sum  of  all  the  verticals  would 
hen  be  equal  to  the  area  of  the  triangle.  Hence,  1-7  (Fig.  104,  g) 

area  of  triangle       Sw  2,       N     .„. 

= p—     -  =  -p.     But  m-7  =  v2=-(i-7)    (^g-   104, 

O 

g)  ;  as  will  now  be  shown.  From  the  similar  triangles  efg  (Fig. 
104,  e)  and  O'm?  (Fig.  104,  d),  ef  :  O'm  : :  f  g  :  in-/.  Also, 
from  the  similar  triangles  e'f'g'  (Fig.  104,  h)  and  O2n7  (Fig. 
104,  g),  e'P  :O2n  : :  f  'g'  =  fg  in-;.  But  ef  =  e'P;  therefore 
m-7  =  n— 7.  Since  the  moment  area  from  which  the  intercepts 
in  Fig.  104,  g  are  taken  is  a  triangle,  it  is  seen  that  m-7  =  n~7  = 

2  ,  2,2  Sw       Sw 

-(1-7).    Now  m-7==v2=j(i-7)  =  =_  ._7==___5  from  which 

3Vol' 
w  =.£__,  (I) 

o 

To  construct  the  deflection  diagram  for  the  beam  fixed  at 
the  right  end  and  supported  at  the  left,  draw  TU  (Fig.  104,  i) 
horizontal,  and  from  this  horizontal  line,  lay  off  ordinates  equal 
to  the  differences  between  the  ordinates  in  Fig.  104,  e  and  Fig. 
104,  h.  Draw  the  polygon  as  shown,  and  trace  a  curve  tangent 
to  this  polygon  at  the  middle  points  of  its  sides,  which  will  be  the 
required  deflection  diagram. 

To  get  the  actual  deflections,  multiply  the  intercepts  in  the 
deflection  diagram  by  the  constant  given  in  equation  (i),  or 
equation  (2),  §  150. 

Simplified  Construction.  A  simpler  construction  for  the  above 
will  now  be  given.  The  bending  moment  diagram,  considering 
the  beam  supported  at  both  ends,  is  FrpGF  (Fig.  104,  b).  To 
draw  the  diagram  for  the  beam  fixed  at  the  right  end  and  sup- 
ported at  the  left,  from  G,  lay  off  GK  =  w  (computed  from 
equation  i),  and  connect  the  points  F  and  K,  completing  the 
fixing  moment  triangle  FGK.  Now  the  fixing  moment  tends  to 


192  MOMENTS,    SHEARS,    AND   DEFLECTIONS.         Chap. 

cause  rotation  in  an  opposite  direction  to  that  caused  by  the  loads 
on  the  beam.  Therefore  the  differences  between  the  ordinates 
in  the  polygon  FrpGF  and  the  triangle  GFK  will  give  the  ordi- 
nates of  the  bending  moment  diagram  for  the  beam  fixed  at 
one  end  and  supported  at  the  other.  This  bending  moment  dia- 
gram is  FrpGKF.  The  moment  at  any  point  along  the  beam  is 
equal  to  the  intercept  in  the  diagram  under  the  point  multiplied 
by  the  pole  distance.  The  point  p  is  the  point  of  contra-flexure, 
i.  e.,  it  is  the  point  where  the  beam  changes  curvature  and  has  zero 
moment.  Since  the  change  in  curvature  of  the  neutral  surface  is 
proportional  to  the  bending  moment,  it  is  evident  that  the  deflec- 
tion diagram  for  the  beam  fixed  at  one  end  and  supported  at  the 
other  may  be  drawn  directly  from  its  bending  moment  diagram. 
To  draw  the  deflection  diagram  by  the  method  suggested,  lay  off 
the  intercepts  between  the  line  FK  (Fig.  104,  b)  and  the  broken 
line  FrpG  on  the  deflection  load  line  (Fig.  104,  j),  noting  that 
intercepts  representing  positive  moments  are  laid  off  in  one  direc- 
tion, and  those  representing  negative  moments,  in  the  opposite 
direction.  From  the  point  7,  draw  the  horizontal  line  7~O3,  and 
with  the  pole  distance  H;?  =  H',  construct  the  funicular  polygon 
(Fig.  104,  i).  Trace  a  curve  tangent  to  this  polygon,  which  will 
give  the  required  deflection  diagram. 

The  latter  method  requires  fewer  constructions,  and  is  more 
accurate  than  the  former. 

It  is  seen  that  the  fixing  moment  at  N  will  decrease  the  reac- 
tion at  M  and  increase  that  at  N  by  an  amount  z,  which  is  the 
force  represented  by  the  distance  between  the  shear  axes  PQ 
(considering  the  beam  supported  at  both  ends)  and  the  shear 
axis  RS  (considering  the  beam  fixed  at  the  right  end  and  sup- 
ported at  the  left). 

To  find  z,  let  Mf  —the  moment  of  the  fixing  couple  at  N. 

Then  Mf  =  wH,  and  z  =— ^-=— —  ,  where  H  =  the  pole  dis- 
tance and  S  =  the  span  of  the  beam.  Compute  z,  and  lay  it  off 
upward  from  P.  Draw  the  shear  axis  RS,  which  is  the  shear  axis 
for  the  beam  fixed  at  the  right  end  and  supported  at  the  left. 


Art.  3. 


BEAM    FIXED    AND    SUPPORTED. 


193 


(b)  Beam  Fixed  and  Supported — Uniform  Load.  It  is  re- 
quired to  draw  the  bending  moment,  shear,  and  deflection  dia- 
grams for  the  beam  MN  (Fig.  105),  fixed  at  the  right  end  and 
supported  at  the  left,  and  loaded  with  a  uniform  load. 


Deflection  Diaqram  -  Simple  Beam 
(e) 


Deflection  Diaqram  -Fixed  and  Supported 


5-6 


FIG.  105.     BEAM  FIXED  AND  SUPPORTED  —  UNIFORM  LOAD. 

Divide  the  uniform  load  into  segments,  and  assume  the  weight 
of  each  segment  as  a  force  acting  through  its  center  of  gravity. 
Apply  the  method  explained  in  the  preceding  article,  and  con- 
struct the  bending  moment,  shear,  and  deflection  diagrams.  Trace 
curves  (not  shown  in  the  diagram)  tangent  to  the  moment  and 
deflection  diagrams,  and  measure  intercepts  between  the  closing 
lines  and  these  curves.  The  bending  moment  diagram  is  shown 
in  Fig.  105,  b;  the  shear  diagram  in  Fig.  105,  c;  the  deflection 
diagram,  considering  the  beam  as  supported  at  both  ends,  in  Fig. 
105,  e;  and  the  deflection  diagram,  for  the  beam  fixed  at  the  right 


194 


BENDING     MOMENTS,    SHEARS,    DEFLECTIONS.  Chap. 


end  and  supported  at  the  left,  in  Fig.  105,  g.    The  intercepts  laid 
bff  in  Fig.  105,  f  are  the  distances  between  the  line  FK  and  the 


-rf- 
^L 


alb  be 

1  j 


Qiagrdms 
(b)  ' 


zz-: 


Shear  Diagram 
(O 


Deflection  Diagram  -  Simple  Beam 
rej 

f 


Deflection  Diagram-Right  fixing  Coup 
(g) 


Deffection  Diagram -Left  Fixinq  Couple 


Deflection  Diagram -Both  Ends  Fixed 

(10 
FIG.  106.     BEAM  FIXED  AT  BOTH  ENDS — CONCENTRATED  LOADS. 


curve  (not  shown),  drawn  tangent  to  the  funicular  polygon.   The 
point  of  contra-flexure  is  at  p  (Fig.  105,  b). 

To  determine  the  actual  deflections,  multiply  the  intercepts  in 


Art.  3.  BEAM  FIXED  AT  BOTH  ENDS.  195 

Fig.  105,  g  by  the  constant  shown  in  equation  (i),  or  equation 

(2),  §  150. 

155.  (3)  Bending  Moment,  Shear,  and  Deflection  Dia- 
grams for  a  Beam  Fixed  at  Both  Ends.  It  is  required  to  draw 
the  bending  moment,  shear,  and  deflection  diagrams  for  the  beam 
MN  (Fig.  106),  fixed  at  both  ends,  and  loaded  with  concentrated 
loads,  as  shown. 

Construct  the  bending  moment,  shear,  and  deflection  dia- 
grams, considering  the  beam  as  supported  at  both  ends,  and 
draw  the  ray  O'm  (Fig.  106,  d),  dividing  the  deflection  load  line 
at  m  into  vx  and  v2. 

» Since  the  beam  is  fixed  at  both  ends,  assume  a  couple  to  act 
at  each  end  whose  moment  is  sufficient  to  make  the  neutral  sur- 
face horizontal  at  these  points.  The  moment  M±,  which  fixes  the 
right  end  of  the  beam,  is  represented  by  the  moment  area  triangle 
FGK  (Fig.  1 06,  b)  ;  and  the  moment  M2,  which  fixes  the  left 
end,  is  represented  by  the  moment  area  triangle  GFL  (see  pre- 
ceding section).  Draw  the  force  polygon  (Fig.  106,  f),  for  the 
moment  triangle  FGK,  and  construct  its  deflection  diagram  (Fig. 
1 06,  g).  Draw  the  dividing  ray  O2n  parallel  to  the  closing 
string,  which  divides  the  load  line  into  v3  and  v4.  Also  draw 
the  force  polygon  (Fig.  106,  h)  for  the  moment  triangle  GFL, 
and  construct  its  deflection  diagram  (Fig.  106,  i).  Draw  the 
dividing  ray  O3n,  which  divides  the  load  line  into  V5  and  ve. 

The  angles  lO'm  and  mO'7  (Fig.  106,  d)  and  their  corre- 
sponding angles  in  the  deflection  diagram  represent  the  deflec- 
tions at  the  ends  of  the  beam,  considered  as  supported  at  both 
ends.  Likewise,  the  angles  iO2n  and  nO2  7  (Fig.  106,  f)  and 
their  corresponding  angles  in  the  deflection  diagram  represent  the 
deflections  at  the  ends  of  the  beam  for  the  right  fixing  couple; 
and  i(Xn  and  nO37  and  their  corresponding  angles  in  the 
deflection  diagram  represent  the  deflections  for  the  left  fixing 
couple.  Since  the  pole  distances  H',  H2  and  H3  are  all  equal,  it 
is  seen  that  vlf  v2,  vs,  v.4,  v5,  and  v6  are  respectively  proportional 
to  these  angles.  To  fix  the  beam  at  the  ends,  i.  e.,  to  make  the 
neutral  surface  horizontal  at  the  ends,  vt  must  equal  V3  +  V5>  and 
v2  must  equal  v4  +  v6. 


196  BENDING     MOMENTS,     SHEARS,     DEFLECTIONS.        Chap.  XVII. 

The  values  of  wl  and  w2  (Fig.  106,  b)  may  be  found,  as 
follows :  As  has  been  shown  in  §  154,  the  length  of  the  deflection 
load  line  1-7  (Fig.  106,  f)  for  the  moment  triangle  FGK  = 

WJL ;   and  this  line  1-7  is  divided  by  the  ray  O2n  into  one-third 


2\' 

and  two-thirds  its  length.    Therefore  v3,  which  equals^  (1-7),= 
W*  ;  and  v4,  which  equals  f  (1-7),=  ^-        In  like  manner, 


considering  the   deflection   load  line   1-7    (Fig.    106,  h)    for  the 
moment  triangle  GFL,  it  may  be  shown  that 


Now,  v,  =  v,  +  vs  ==          +      p.  ,  (2) 

and       '  v,  =  v4  +  v.=|£  +  S£.-  (3) 

Solving  equations  (2)  and  (3)  for  wx  and  w,,  we  have 

2\' 
wi=     --C^Vi  —  v2),  (4) 


and  w2  =  --  (2v2  —  vj.  (5) 

O 

To  get  the  bending  moment  at  any  point  for  the  beam  fixed 
at  both  ends,  compute  wl  and  w2  from  the  above  formulae,  and 
lay  them  off  downward  from  F  and  G  (Fig.  106,  b),  respectively. 
Join  the  points  L  and  K.  Then  the  moment  area  FGKL  repre- 
sents the  bending  moment  of  the  fixing  couples  at  both  ends  of 
the  beam.  Since  the  polygon  Fprp'GF  represents  the  bending 
moment,  considering  the  beam  supported  at  both  ends,  it  is  seen 
that  the  bending  moment  for  the  beam  fixed  at  both  ends  is  repre- 
sented by  the  moment  area  Fprp'GKLF.  The  moment  at  any 
point  is  equal  to  the  intercept  in  this  diagram  multiplied  by  the 
pole  distance  H. 

The  points  p  and  p'  are  points  of  contra-flexure. 


Art.  3.  BEAM    FIXED    AT    BOTH    ENDS.  197 

To  draw  the  deflection  diagram  (Fig.  106,  k)  for  the  beam 
fixed  at  both  ends,  construct  the  deflection  load  line  (Fig.  106,  j) 
by  laying  off  on  this  load  line  distances  equal  to  the  intercepts 
between  the  line  LK  (Fig.  106,  b)  and  the  broken  line  Fprp'G. 
The  intercepts  6-1  and  1-2  are  laid  off  in  an  upward  direction, 
while  2-3,  3-4,  4-5,  and  5-6  are  laid  off  in  a  downward  direction. 
Farom  the  point  i,  with  the  pole  distance  H4  —  H',  draw  the  hori- 
zontal line  i-O4.  Construct  the  funicular  polygon  (Fig.  106, 
k),  and  draw  a  curve  tangent  to  the  polygon  at  the  middle 
points  of  its  sides,  whictuwill  give  the  required  deflection  dia- 
gram. It  is  seen  that  the  neutral  surface  is  horizontal  at  the  ends 
of  the  beam.  To  obtain  the  actual  deflections,  multiply  the  inter- 
cepts in  Fig.  1 06,  k  by  the  constant  shown  in  equation  (i),  or 
equation  (2),  §  150. 

The  deflection  diagram  might  also  have  been  drawn  by  sub- 
tracting from  the  intercepts  in  Fig.  106,  e  the  sum  of  the  corre- 
sponding intercepts  in  Fig.  106,  g  and  Fig.  106,  i;  and  laying 
off  their  differences  from  the  horizontal  line  TU. 

The  former  method  is  preferable,  and  is  subject  to  less  error 
than  the  latter.  When  the  former  described  method  is  used,  the 
diagrams  shown  in  Fig.  106,  f,  Fig.  106,  g,  Fig.  106,  h,  and  Fig. 
106,  i  need  not  be  drawn  except  to  prove  the  constructions. 

To  construct  the  shear  diagram,  it  is  necessary  to  find  the 
effect  upon  the  reactions  due  to  the  fixing  moment  at  each  end. 

Let  zl  =  decrease  in  the  reaction  at  M  due  to  the  fixing 
moment  at  N, 

and  z2  =  'decrease  in  the  reaction  at  N  due  to  the  fixing  moment 
at  M. 

Then  z0 —  z:  =  distance  (to  scale  of  forces)  between  shear  axes 
PQ  and  RS  (Fig.  106,  c). 

Now         Zl  =  ^,  and  za=^?  (see  §  154  for  proof). 

o  ^ 

TT 

Therefore,  z2  —  zi=-g-  (w»  —  w,).  (6) 

If  z,  is  greater  than  z,   (as  in  this  problem),  z, —  zt  is  to  be 


198  BENDING     MOMENTS,    SHEARS,     DEFLECTIONS.        Chap.  XVII. 


measured  upward  from  PQ  (the  shear  axis  for  the  beam  sup- 
ported at  both  ends)  ;  and  if  zt  is  greater  than  z2,  then  z2  —  z±  is 
to  be  measured  downward  from  PQ. 

Draw  the  shear  diagram  and  shear  axis  PQ,  considering  the 
beam  as  supported  at  both  ends,  lay  off  z2  —  zx  upward  from  PQ, 
and  draw  the  shear  axis  RS,  which  is  the  shear  axis  for  the  beam 
fixed  at  both  ends. 


BEAM  &  LOADING 

MAX.  MOMENT 

MAX-  DEFLECTION 

IP                         w 

-PL 

PL3 

HI 

W=wL 

NA/L 

WL3 

IP 

2 

PL3 

W=wL 

48  El 

CM/I  3 

f  \ 

,  WL 

D/V  L. 

IP    • 

h  8 
.5PL,     3  PL 

384  El 
0.0093  PL3 

t                      H 

h  32           16 

El 

r~21 

,  PL  ,      PL 

0-0054  WL 
El 

PL3 

a,      W=wL      p 

f  8            8 

192  El 

m          n           H 

,  WL          WL 

WL 

^i                                          Kl 

f  24    '       12 

384  El 

FIG.  107.     FORMULAE  FOR  MAXIMUM  MOMENTS  AND  DEFLECTIONS. 

156.  Algebraic  Formulae.  In  Fig.  107  are  given  several 
algebraic  formulae  for  determining  the  maximum  bending  mo- 
ments and  maximum  deflections  in  beams  for  some  of  the  simpler 
forms  of  loading. 


CHAPTER  XVIII. 


MAXIMUM  BENDING  MOMENTS    AND    SHEARS    IN    BEAMS    FOK 
MOWNG   LOADS. 


This  chapter  will  treat  of  the  determination  of  the  maximum 
bending  moment  and  maximum  shear  at  any  point  in  a  simple 
beam  loaded  with  moving  loads ;  and  of  the  position  of  the  mov- 
ing loads  for  a  maximum  bending  moment  or  maximum  shear. 


M 


Uniform  load  of  p  Ibs.per  lin- ft-of  span 


Maximum  Moment  Diagram 
FIG.  108.     SIMPLE  BEAM — UNIFORM  LOAD. 

157.  Beam  Loaded  with  a  Uniform  Load,  (a)  Maximum 
Bending  Moment.  It  is  required  to  find  the  maximum  bending 
moment  at  any  point  O  (Fig.  108)  of  a  beam  loaded  with  a 
uniform  load.  First  assume  the  beam  to  be  unloaded,  and  then 
assume  a  uniform  load  to  move  onto  the  beam.  The  bending 
moment  at  every  point  along  the  beam  is  increased  with  each  addi- 
tion of  the  uniform  load  until  it  reaches  its  maximum  value  at 
every  point  when  the  beam  is  fully  loaded.  It  has  been  shown  in 
§  147,  b  that  the  bending  moment  diagram  for  a  beam  fully 
loaded  with  a  uniform  load  is  a  parabola  with  a  maximum  ordi- 
nate  at  the  center  of  the  beam  equal  to  JpL2.  The  maximum 
bending  moment  diagram  for  the  beam  MN  is  shown  in  Fig.  108. 

199 


200 


MAXIMUM    BENDING    MOMENTS   AND    SHEARS.       Chap.  XVIII. 


The  bending  moment  at  any  point  O  (Fig.  108),  at  a  distance 
x  from  the  center  of  the  beam  is 

iv/r         pL"         px"°  /   N 

M=— o-— -  — -  (i) 

O  2.    ' 


The  above  equation  may  be  written 


(2) 


This  equation  expressed  in  words  is:  The  bending  moment  at 
any  point  in  a  beam  uniformly  loaded  is  equal  to  one-half  the  load 
per  foot  multiplied  by  the  product  of  the  two  segments  into  which 
the  beam  is  divided  by  the  given  point. 

The  maximum  bending  moment  at  any  point  along  a  beam 
uniformly  loaded  may  be  readily  found  from  equation  (2). 

(b)  Maximum  Shear.  It  is  required  to  find  the  maximum 
shear  at  any  point  O  of  the  beam  MN  (Fig.  109),  loaded  with  a 
uniform  load. 


Uniform  load  of  p  Ibs.per  I'm. ft 


Maximum  Shear  Diagram 


FIG.  109.     SIMPLE  BEAM — UNIFORM  LOAD. 

It  has  been  shown  in  §   147,  b  that  the  equation  express- 
ing the  shear   in  a  beam  fully  loaded  with  a  uniform  load  is 

/L          \ 
S  =  p  f x  1     where  p  is  the  uniform  load  per  foot,  L  the 

length  of  the  beam,  and  x  the  distance  from  the  left  end  of  the 


SIMPLE    BEAM — UNIFORM    LOAD.  201 

beam  to  the  point  where  the  shear  is  to  be  determined.    It  is  seen 
from  this  equation  that  when  the  beam  is  fully  loaded  the  shear 

has  its  maximum  positive  value  -f-          when  x  =  o;   that  it  has 

2 

its  maximum  negative  value wnen  x  =  L ;    and  is  zero 

^  2 

when  x  = 

2 

Now  it  is  seen  that  the  load  to  the  left  of  any  point  O  (Fig. 
109)  decreases  the  positive  shear  at  that  point  by  the  amount  of 
the  load,  while  it  increases  the  left  reaction  by  a  less  amount. 
Since  the  shear  is  equal  to  the  reaction  minus  the  load  to  the  left, 
it  will  have  its  maximum  value  at  any  point  when  there  is  no  load 
to  the  left  of  the  point.  Therefore,  to  get  a  maximum  positive 
shear  at  any  point  in  a  beam  due  to  a  uniform  load,  the  segment 
to  the  left  of  the  point  should  be  unloaded  and  that  to  the  right 
fully  loaded. 

The  equation  for  a  maximum  positive  shear  is 

R        PlL-xHk^L»=   P    (L-x)».  (3) 

2L  2L 

This  is  the  equation  of  the  parabola  CB,  which  has  its  vertex 
at  the  right  end  of  the  beam.    The  maximum  ordinate  is  at  x  =  o, 

the  left  end  of  the  beam,  and  is  equal  to  _ — 

2 

To  get  the  maximum  negative  shear  at  any  point,  the  portion 
to  the  left  of  the  point  should  be  fully  loaded  and  that  to  the 
right  unloaded.  The  equation  expressing  the  maximum  negative 
shear  is 

S  —  R.— -^  (4) 

2L 

which  is  the  equation  of  the  parabola  AD. 

158.     Beam  Loaded  with  a  Single  Concentrated  Load,     (a) 

Maximum  Bending  Moment.     Let  the  beam  MN  (Fig.  no)  be 


202 


MAXIMUM    BEXDiXG    MOMEXTS   AND    S1IEAKS.       Chap.  XTIII. 


loaded  with  a  single  moving  load  P.  The  maximum  bending 
moment  at  any  point  O  occurs  when  the  load  is  at  that  point ;  for 
a  movement  of  the  load  to  either  side  of  the  point  decreases  the 


M 

p 

N 

|R,      |-x      55 

k 

4..                                  ...2-    . 

}' 

(a)   Maximum  Moment  Diagram 


(b)  Maximum  Shear  Diagram 
FIG.   110.     SIMPLE  BEAM — CONCENTKAIED  MOVING  LOAD. 


opposite  reaction  and  hence  the  bending  moment.  The  equation 
expressing  the  maximum  bending  moment  at  any  point  due  to  a 
single  moving  load  is 


(5) 


This  is  the  equation  of  the  parabola   (Fi-g.   no,  a),  which  has 


PL 


when  x  =  o. 


its  maximum  ordinate  equal  to 

4 

2p 
If  -j—  is  substituted  for  P  in  equation  (i),  it  is  seen  that  we 

have  equation  (5).  Therefore,  the  bending  moment  due  to  a  sin- 
gle concentrated  load  is  the  same  as  for  twice  that  load  uniformly 
distributed  over  the  beam. 

(b)     Maximum  Shear.     To  get  the  maximum  positive  shear 


SIMPLE  BEAM-CONCENTRATED  MOVING  LOADS.  203 

at  any  point  O  (Fig.  no)  due  to  a  single  moving  load,  the  load 
should  be  placed  an  infinitesimal  distance  to  the  right  of  the 
point.,  (For  all  practical  purposes  it  may  be  considered  at  the 
point.)  The  maximum  shear  may  be  expressed  by  the  equation 


which  is  the  equation  of  the  straight  line  CB.    The  positive  shear 

L 
is  a  maximum  when  x  =  —  ,  i.  e.,  at  the  left  end  of  the  beam. 

The  maximum  negative  shear  occurs  when  the  load  is  just  to 
the  left  of  the  point,  and  may  be  expressed  by  the  equation 


which  is  the  equation  of  the  straight  line  AD. 

159.     Beam  Loaded  with  Concentrated  Moving  Loads,    (a) 

Position  for  Maximum  Moment.  Let  MN  (Fig.  in)  be  a  beam 
loaded  with  concentrated  moving  loads  at  fixed  distances  apart. 
(For  simplicity,  three  loads  are  here  taken,  although  any  number 


*  -d5 


w 

M  v  y  ooiv  y     ffl 


R, 


L-x 


FIG.  111.     POSITION  FOR  MAXIMUM  MOMENT — MOVING  LOADS. 

might  have  been  considered.)  It  is  required  to  find  the  posi- 
tion of  the  loads  for  a  maximum  bending  moment  in  the  beam, 
together  with  the  value  of  this  moment. 

The  determination  of  the  position  of  any  number  of  moving 
loads  for  a  maximum  bending  moment  at  any  point  in  a  beam  will 
be  treated  in  §  208. 

Let  x  be  the  distance  of  one  of  the  loads  P.,  from  the  left  end 


204  MAXIMUM    BENDING    MOMENTS    AND    SHEARS.       Chap.  XVIII. 

of  the  beam  when  the  loads  are  so  placed  that  they  produce  a 
maximum  moment  under  P2.  Now  if  the  bending  moment  is 
found,  and  its  first  derivative  placed  equal  to  zero,  the  position  of 
the  loads  for  a  maximum  bending  moment  may  be  determined. 
To  determine  the  bending  moment,  first  find  the  reaction  Rx  by 
taking  moments  about  the  right  end  of  the  beam.  Thus, 

R  ^P!  (L  — x  +  a)  +P2  (L  — x)  +  P3  (L  — x  — b) 

_  (P1  +  P2  +  P3)  (L  — x)  +  P,a  —  P3b 
= _ ^  \&) 

L, 

and  the  bending  moment  under  P2  is 

M  =  RlX—  Pia 
x  (Pj  +  p^-f-p,,)  (L  — x)  +x  (P,a—  P.-b) 

TT  Pia-    (9) 

Differentiating  equation  (9)  and  placing  it  equal  to  zero,  we  have 

dM        (pt -|- p0  +  p8)  (L  — 2x)  +P±a— Pnb 

— r  = = : = =o.        (10) 

dx  L 

Solving  equation  (10)  for  x,  we  have 

L  pia_pob 

x==T  +  g(Pi  +  pa  +  P.)-  (II) 

The  position  of  the  wheel  P2  for  a  maximum  bending  moment 
in  the  beam  is  determined  by  equation  ( 1 1 ) .  For,  Pxa  —  P3b  =  the 


i3 
moment  of  the  loads  on  the  beam  about  P2,  and  p        p        p   = 

*1  ~T.  *  S  ~l      ^3 

distance  from  P2  to  the  center  of  gravity  of  all  the  loads ;   hence 
x  = 1 —  (distance  from  P2  to  c.  g.  of  all  the  loads). 

Therefore,  the  criterion  for  a  maximum  bending  moment  under 
the  load  P2  is :  The  load  P2  must  be  as  far  from  one  end  of  the 
beam  as  the  center  of  gravity  of  all  the  loads  is  from  the  other  end. 


SIMPLE  BEAM-COXCEXTRATED  MOVING  LOADS.  205 

Since  P2  is  any  one  of  the  moving  loads,  it  is  seen  that  theo- 
retically  this  criterion  must  be  applied  to,  and  the  bending  mo- 
ment found  for,  each  of  the  loads ;  and  the  greatest  value  taken 
as  the  maximum  bending  moment.  However,  it  may  often  be 
determined  by  inspection  which  load  will  give  a  maximum  mo- 
ment. If  some  of  the  loads  are  heavier  than  others,  the  maximum 
moment  will  occur  under  one  of  the  heavier  loads. 

If  x  had  been  measured  from  the  center  of  the  beam  instead 
of  from  the  ends,  the  following  equivalent  criterion  for  the  maxi- 
mum bending  moment  would  have  been  found :  For  a  maximum 
bending  moment  under  the  load  P2,  this  load  must  be  as  far  to 
one  side  of  the  center  of  the  beam  as  the  center  of  gravity  of  all 
the  loads  is  to  the  other  side  of  the  center. 

The  methods  employed  for  determining  the  bending  moment 
for  fixed  loads  may  be  applied  to  moving  loads  as  soon  as  their 
position  for  a  maximum  bending  moment  has  been  found. 


M 


(9  ^  Q 


Rl     ?~1 

.~_fct£ 


L 

FIG.  112.     POSITION  FOR  MAXIMUM  MOMKNT — Two  EQUAL  LOADS. 


A  special  case  of  the  above,  which  often  occurs,  is  that  of  a 
beam  loaded  with  two  equal  loads  at  a  fixed  distance  a  apart 
(Fig.  112).  Applying  the  criterion  for  a  maximum  moment  to 

this  case,  it  is  seen  that  x  (measured  from  the  end)  =  —  4.  — , 

2         4 

or  if  measured  from  the  center  =  — . 


Placing  one  of  the  loads  P  at  a  distance  —from  the  center  of  the 

4 

beam,  and  taking  moments  about  the  left  end   (noting  that  the 
loads  are  equal),  we  have 


206  MAXIMUM    BENDING    MOMENTS    AND    SHEARS.       Chap.  XVIII. 


L 

and  the  maximum  moment  M  is 


= •  (12) 

2L 

This  equation  is  true  for  the  ordinary  values  of  a  and  L,  but 
does  not  give  a  maximum  bending  moment  when  a  >  O.586L. 
When  a  >  0.586!.,  one  of  the  wheels  placed  at  the  center  of  the 
beam  (the  other  being  then  off  of  the  beam)  will  give  a  maximum 
bending  moment,  as  will  now  be  shown. 

Assume  that  one  of  the  two  equal  loads  P  is  placed  at  the 
center  of  the  beam,  and  that  the  distance  a  between  the  loads  is 
greater  than  0.5!..  The  maximum  bending  moment  is  then  equal 

PL 

to Equating  this   to  the   value   found   for   the   maximum 

4 
bending  moment  in  equation  (12'),  we  have 


4  2L 

Solving  for  a,  we  have 

a  =  o.586L.  (13) 

Therefore,  when  a  =  O.586L,  the  moment  clue  to  a  single  load 
P  at  the  center  of  the  beam  is  the  same  as  the  moment  given  by 
placing  the  loads  according  to  the  criterion  for  maximum  mo- 
ments. When  a  >  O-586L,  the  maximum  bending  moment  for 
two  equal  loads  at  a  fixed  distance  apart  is  given  by  placing  one 
of  the  loads  at  the  center  of  the  beam ;  and  the  criterion  for  a 
maximum  bending  moment  does  not  apply. 

If  the  two  loads  are  unequal,  the  maximum  moment  will 
always  occur  under  the  heavier  load. 


SIMPLE  BEAM-COXCEXTRATED  MOVING  LOADS.  207 

(b).  Position  for  Maximum  Shear.  For  two  equal  loads, 
the  maximum  end  shear  will  occur  when  both  loads  are  on  the 
span  and  when  one  of  the  loads  is  at  an  infinitesimal  distance 
from  the  end  of  the  beam ;  since  the  reaction  will  be  a  maximum 
for  this  condition.  For  a  maximum  shear  at  any  point  along  a 
beam  due  to  two  equal  loads,  one  of  the  loads  must  be  at  the  point. 

For  two  unequal  loads,  the  maximum  end  shear  will  occur 
when  both  loads  are  on  the  span  and  when  the  heavier  load  is 
at  an  infinitesimal  distance  from  the  end.  For  a  maximum  shear 
at  any  point  along  the  beam  due  to  two  unequal  loads,  the  heavier 
load  must  be  at  the  point. 

For  a  maximum  shear  at  any  point  along  a  beam  due  to  any 
number  of  loads,  one  of  the  loads  must  be  at  the  point.  The 
criterion  for  determining  which  load  placed  at  the  point  will  give 
a  maximum  shear  will  now  be  determined. 

Let  MN  (Fig.  113)  be  a  beam  loaded  with  any  number  (in  this 
case  four)  concentrated  loads,  and  let  O  be  any  point  along  the 
beam  whose  distance  from  the  left  end  is  x.  Also  let  2P '=  the 
sum  of  all  the  loads  on  the  beam  when  there  is  a  maximum  shear. 
It  is  required  to  determine  which  load  placed  at  O  will  give  a 
maximum  shear  at  that  point 

i*    IP* 

M 


IR        °   u_q_i>_  '  _c  tR 

b'::::r^::-::::::c:-l:::v.:::::::d' 

FIG.  113.     POSITION  FOR  MAXIMUM   SHEAR — MOVING  LOADS. 

When  Px  is  placed  at  O,  the  shear  Si  at  O  is  equal  to  the  left 
reaction  R1?  i.  e., 

p    /L x)  4- P.,  (L x a) 


P,  [L  —  x—  (a  +  b)]+P4[L  —  x--(a  +  b 

'  — 

2P(L  — x)— P2a— P3(a  +  b)— P4(a+b 


208  MAXIMUM   SHEAR.  Chap.  XVIII. 

When  Po  is  at  O,  the  shear  S2  at  O  is 

P1(L-x  +  a)+P2(L-x) 
b2  =  Kx  —  ^  -  —  — 

LP3(L-x-b)+P4[L-x-(b  +  c)1 

L  l 

_  3P  (L  —  x)  +Pia  —  Psb  —  P4  (b  +  c)—  PaL 
~L~~ 

Subtracting  S2  from  Sj,  we  get  the  difference  in  the  shear  for  the 
two  cases,  or 


, 

From  the  above  equation,  it  is  seen  that  St  will  be  greater  than 
S2  if  PjL  >  2Pa,  i.  e.,  if 


The  above  equation  expressed  in  words  is  :  The  maximum  pos- 
itive shear  at  any  point  along  a  beam  occurs  when  the  foremost 

.  PiL  P,L 

load  is  at  the  point  if-  —is  greater  than  2P.    //  -  is  less  than 
a  a 

SPj,  the  greatest  shear  will  occur  when  some  succeeding  load 
(usually  the  second)  is  at  the  point. 

The  methods  employed  to  determine  the  shear  due  to  fixed 
loads  may  be  applied  to  moving  loads  as  soon  as  their  position 
for  a  maximum  shear  has  been  determined. 


PART  IV. 
BRIDGES. 

CHAPTER  XIX. 

TYPES  OF  BKTDGE  TRUSSES. 

Bridge  trusses  are  comparatively  recent  structures,  the  ancient 
bridges  being  pile  trestles  or  arches.  Somewhat  later,  a  combina- 
tion of  arch  and  truss  was  used,  although  the  principles  govern- 
ing the  design  were  not  understood.  It  was  not  until  1847  tnat 
the  stresses  in  bridge  trusses  were  fully  analyzed,  although  trusses 
were  constructed  according  to  the  judgment  of  the  builder  before 
this  date.  In  1847,  Squire  Whipple  issued  a  book  upon  bridge 
building,  and  he  was  the  first  to  correctly  analyze  the  stresses 
in  a  truss.  Soon  afterward,  the  solution  of  stresses  became  very 
generally  understood,  wooden  trusses  were  discarded  for  iron 
ones,  and  still  later,  steel  replaced  iron  as  a  bridge-truss  mate- 
rial. From  this  time,  the  development  of  bridge  building  was 
very  rapid,  culminating  in  its  present  high  state  of  efficiency. 

160.  Through  and  Deck  Bridges.  Bridges  may  be  grouped 
into  two  general  classes,  viz. :  through  bridges  and  deck  bridges. 

A  through  bridge  is  one  in  which  the  floor  is  supported  at,  or 
near,  the  plane  of  the  lower  chords  of  the  trusses  (see  Fig. 
114,  e).  The  traffic  moves  through  the  space  between  the  two 
trusses.  Except  in  the  case  of  a  pony  truss  (one  in  which  there 
is  no  overhead  bracing),  a  system  of  overhead  lateral  bracing  is 
used. 

209 


210 


TYPES   OF   BRIDGE   TRUSSES. 


Chap. 


A  deck  bridge  is  one  in  which  the  floor  is  supported  directly 
upon  the  upper  chords  of  the  trusses.  In  this  type,  the  trusses 
are  below  the  floor  (see  Fig.  114,  d). 

161.  Types  of  Bridge  Trusses.  In  Fig.  114  are  shown 
several  types  of  bridge  trusses  that  have  been  very  generally 
used. 


A7WW\. 


(a)  Warren 


(b)  Howe 


(c)  Pratt 


(d)  Baltimore  ( Deck) 


(e)  Baltimore  (Thru) 


(f)  Whipple 


(g)  Camels  Back 


(h)  Parabolic  Bowstring 


(i)  Parabolic  Bowstring 


ij)  Petit 
FIG.  114.     TYPES  OF  BRIDGE  TRUSSES. 

Fig.  114,  a  shows  a  Warren  truss.  This  truss  is  still  used  for 
short  spans,  but  has  the  disadvantage  that  the  intermediate  web 
members  are  subjected  to  reversals  of  stress. 

Fig.   114,  b  shows  a  Howe  truss.     This  truss  was  in  favor 


TYPES    OF    BRIDGE    TRUSSES.  '2ll 

when  wood  was  extensively  used  as  a  building  material  for 
trusses,  but  is  little  used  at  present.  It  has  the  disadvantage  of 
having  long  compression  web  members. 

Fig.  114,  c  shows  a  Pratt  truss.  This  form  of  truss  is  exten- 
sively used,  both  for  highway  and  railroad  bridges,  up  to  about  a 
2OO-foot  span.  It  is  economical  and  permits  of  good  details. 

Fig.  114,  d  shows  a  Baltimore  deck  truss,  and  Fig.  114,  e,  .a 
Baltimore  through  truss.  These  trusses  are  used  for  compara- 
tively long  spans,  and  have  short  compression  members. 

Fig.  114,  f  shows  a  Whipple  truss.  This  truss  was  quite  exten- 
sively used,  but  is  now  seldom  employed.  It  is  a  double  intersec- 
tion truss,  and  has  a  redundancy  of  web  members.  The  stresses 
are  indeterminate  by  ordinary  graphic  methods. 

Fig.  114,  g  shows  a  Camels-Back  truss.  This  truss  is  used 
both  for  short  and  long  spans. 

Fig.  114,  h  and  Fig.  114,  i  show  Parabolic  Bowstring  trusses. 
The  upper  chord  panel  points  are  on  the  arc  of  a  parabola.  A 
great  disadvantage  of  these  types  is  that  the  upper  chord  changes 
direction  at  each  panel  point  and  that  the  web  members  change 
both  their  angle  of  inclination  and  length  at  the  panel  points. 

This  type  is  sometimes  modified  by  placing  the  panel  points 
on  the  arc  of  a  circle. 

Fig.  114,  j  shows  a  Petit  truss.  This  truss  is  quite  exten- 
sively used  for  long  spans,  and  is  economical. 

162.  Members  of  a  Truss.  The  general  arrangement  of 
members  is  given  in  the  through  Pratt  railroad  truss  shown  in 
Fig.  115.  The  arrangement  of  members  in  the  various  types  of 
trusses  is  somewhat  similar  to  that  shown.  In  this  truss,  the  ten- 
sion members  are  shown  by  light  lines,  and  the  compression  mem- 
bers by  heavy  lines. 

Main  Trusses.  Each  truss  consists  of  a  top  chord,  a  bottom 
chord,  an  end  post,  and  web  members.  The  web  members  may 
be  further  subdivided  into  hip-verticals,  intermediate  posts,  and 
diagonals.  The  diagonals  may  be  divided  into  main  members  and 
counters,  the  main  members  being  those  stressed  under  a  dead 
load,  and  the  counters  those  stressed  only  under  a  live  load.  In 
Fig.  115,  LoUj  is  an  end  post;  UYUo,  a  panel  length  of  the  upper 


212 


TYPES    OF    BRIDGE    TRUSSES. 


Chap.  XIX. 


chord;    L^,  a  panel  length  of  the  lower  chord;    U^,  a  hip 
vertical ;  UjL,,  and  U2L3,  main  diagonals  ;  and  L2U3,  a  counter. 
Lateral  Bracing.     The   bracing   in   the   plane   of   the   upper 

Top  Lateral  Strut 
op  Lateral  Ties 


Stringers 
Pedestal 


-Intermediate  Post 
Floorbeams 
Bottom  Lateral  Ties 

THROUGH  PRATT  TRUSS 

PIG.  115.     ARRANGEMENT  OF  TRUSS  MEMBERS. 


chord  (Fig.  115)  is  called  the  top  lateral  bracing;  and  that  in 
the  plane  of  the  lower  chord,  the  bottom  lateral  bracing.  The 
members  of  the  lateral  systems  are  stressed  by  wind  loads  and  by 
the  vibrations  due  to  live  loads.  The  top  lateral  system  is  com- 
posed of  top  lateral  struts  and  ties.  The  floorbeams  act  as  the 
struts  in  the  lower  lateral  system. 

Portals.  In  through  bridges,  the  trusses  are  held  in  position 
and  the  bridge  made  rigid  by  a  system  of  bracing  in  the  planes 
of  the  end  posts.  This  system  of  bracing  is  called  the  portal 
bracing,  or  portal. 


MEMBERS  OF  A  TRUSS.  213 

Knee-braces  and  Sway  Bracing.  The  braces  connecting  the 
top  lateral  struts  and  intermediate  posts  (see  Fig.  115),  in  the 
plane  of  the  intermediate  posts,  are  called  knee-braces.  When 
greater  rigidity  is  required,  a  system  of  bracing  somewhat  similar 
to  the  portal  bracing  is  used  instead  of  the  knee-braces.  This 
bracing  is  called  the  sway  bracing.  The  top  lateral  strut  is  also 
the  top  strut  of  the  sway  bracing.  Knee-braces  and  sway  bracing 
are  often  omitted  on  small  span  highway  bridges. 

Floor  System.  The  floor  systems  of  ordinary  highway  bridges 
differ  considerably  from  those  of  railroad  bridges.  Both  types, 
however,  have  cross-beams  running  from  one  hip  vertical  or  inter- 
mediate post  to  the  opposite  one.  These  beams  are  called  floor- 
beams.  The  beams  at  the  ends  of  the  bridge  are  called  the  end 
floorbeams,  and  those  at  the  intermediate  posts,  intermediate  floor- 
beams.  The  end  floorbeams  are  usually  omitted  in  highway 
bridges,  and  an  end  strut,  or  joist  raiser,  is  substituted. 

In  railroad  bridges,  there  are  beams  which  run  parallel  to  the 
chords  and  are  connected  at  their  ends  to  the  floorbeams.  These 
beams  are  called  stringers. 

In  highway  bridges,  there  are  several  lines  of  beams  which 
run  parallel  to  the  chords  and  which  rest  upon  the  floorbeams. 
These  beams  are  called  joists. 

In  railroad  bridges,  the  ties,  which  support  the  rails,  rest 
directly  upon  the  stringers ;  and  in  highway  bridges,  the  floor 
surface  is  supported  directly  by  the  joists. 

Pedestals.  The  supports  for  the  ends  of  the  trusses  are  called 
pedestals.  For  spans  over  about  70  feet,  the  pedestals  at  one  end 
of  the  bridge  are  provided  with  rollers,  to  allow  for  expansion 
and  contraction. 

Connections.  The  members  of  the  truss  may  be  either  riveted 
together  or  connected  by  pins.  In  the  former  case,  the  truss  is 
said  to  be  a  riveted  truss,  and  in  the  latter,  a  pin-connected  truss. 
Riveted  trusses  are  often  used  for  short  spans  and  are  very  rigid. 
Pin-connected  trusses  are  easy  to  erect,  and  are  used  for  both 
short  and  long  spans. 


CHAPTER  XX. 
LOADS. 

The  loads  for  which  a  bridge  must  be  designed  may  be  classi- 
fied, as  follows:  dead  load,  live  load,  and  wind  load;  and  these 
loads  will  be  discussed  in  the  following  three  articles. 

ART.  i.     DEAD  LOAD. 

The  dead  load  is  the  weight  of  the  entire  bridge,  and  includes 
the  weights  of  the  trusses,  bracing,  floor,  etc.  It  is,  of  course, 
necessary  to  determine  the  weight  of  the  trusses  before  the  dead 
load  stresses  in  them  may  be  determined;  therefore  an  assump- 
tion must  be  made  as  to  the  dead  load.  If  the  weights  of  similar 
bridges  are  available,  then  these  weights  may  be  used  to  determine 
the  dead  load ;  but  it  is  customary  to  use  formulae  for  finding  the 
approximate  dead  load.  It  should  be  borne  in  mind  that  the  dead 
load  for  a  single-track  bridge  is  carried  by  two  trusses,  each  sup- 
porting one-half  the  total  load. 

163.  Weights  of  Highway  Bridges.  The  total  dead  load 
per  linear  foot  of  span  for  a  highway  bridge  not  carrying  inter- 
urban  cars  may  be  very  closely  approximated  by  the  formula* 

w=  140+  !2b  +  o.2bL  —  0.4  L,  (i) 

where  w  =  weight  of  bridge  in  pounds  per  linear  foot, 

b==  width  of  bridge  in  feet  (including  sidewalks,  if  any), 
L  =  span  of  bridge  in  feet. 


*Merriman  and  Jacoby's  "Boofs  and  Bridges,"  Part  II. 

214 


4-rt.l.  DEAD    LOAD.  215 

The   weight  of   a   heavy   interurban   riveted   bridge   may   be 
closely  approximated  by  the  formulaf 


),  (2) 

where, 

w  =  weight  of  bridge  in  pounds  per  linear  foot, 
b  =  width  of  roadway  (including  sidewalks), 
L  =  span  of  bridge  in  feet. 
164.     Weights  of  Railroad  Bridges.     The  weights  of  rail- 

road bridge  trusses  per  linear  foot  of  span  are  given  very  closely 

by  the  following  formulae  :  J 


Eso,     w=     (650  +  ;L),  (3) 

E4o,     w  =  J(65o  +  7L),  (4) 

E3o,     w  =  }(6so  +  7L).  (5) 


The  above  formulae  do  not  include  the  weights  of  the  ties  and 
rails,  which  may  be  assumed  at  400  pounds  per  linear  foot  of 
track.  If  solid  steel  floors  are  used,  the  weight  of  the  track 
should  be  taken  at  700  pounds  per  linear  foot. 

£50,  £40,  and  £30  refer  to  the  live  load  for  which  the 
bridge  is  designed  ;  as  given  by  Theodore  Cooper  in  his  "General 
Specifications  for  Steel  Railroad  Bridges  and  Viaducts."  This 
live  loading  will  be  explained  in  §  167  (b). 

These  formulas  give  the  weights  of  single-track  spans.  Double- 
track  spans  are  about  95  per  cent  heavier. 


ART.  2.     LIVE  LOAD. 

The  live  load  consists  of  the  traffic  moving  across  the  bridge. 
For  highway  bridges,  the  live  load  consists  of  vehicles,  foot  pas- 
sengers, and  interurban  cars  ;  and  for  railroad  bridges  it  consists 

tE.  S.  Shaw. 

$F.  E.  Turneaure. 


216  LOADS.  Chap.  XX. 

of  trains.  The  standard  highway  bridge  specifications  of  Theo- 
dore Cooper  and  J.  A.  L.  Waddell  give  the  live  loads  to  be  used 
for  different  classes  of  highway  bridges,  and  are  recommended  by 
the  writer  for  general  use. 

165.  Live  Load  for  Light  Highway  Bridges.  The  live 
load  for  highway  bridges  is  usually  specified  in  pounds  per  square 
foot  of  floor  surface.  The  weights  given  in  the  following  table 
are  recommended  for  use  when  the  bridge  does  not  carry  inter- 
urban  traffic. 


LIVE  LOADS  FOR  HIGHWAY  BRIDGE  TRUSSES. 

Spans     up  to  100  feet 100  Ibs.  per  sq.  ft. 

100  to  125    " 95    "       "      " 

125  to  150    "     90    "       "      " 

150    tO    200     "      85     "          "        "       " 

"        over       200    "    80    "       "      "     " 

In  some  states,  the  law  requires  that  highway  bridges  be 
designed  for  a  live  load  of  100  pounds  per  square  foot  of  floor 
surface,  but  the  law  as  usually  stated  is  defective  in  that  the 
allowable  unit  stresses  are  not  given. 

The  floor  systems  of  light  highway  bridges  should  be  designed 
to  carry  a  live  load  of  100  pounds  per  square  foot  of  floor  space, 
and  to  be  of  sufficient  strength  to  carry  a  heavy  traction  engine. 
The  total  load  is  obtained  by  multiplying  the  weight  per  square 
.foot  by  the  clear  width  of  the  roadway  and  sidewalks ;  and  this 
load  is  to  be  so  placed  as  to  give  the  maximum  stress  in  each  truss 
membeV. 

166.  Live  Load  for  Interurban  Bridges.    For  the  live  load 
on  interurban  bridges,  the   student  is   referred  to  the  highway 
bridge  specifications  of  Theodore  Cooper  and  J.  A.  L.  Waddell. 

167.  Live  Load  for  Railroad  Bridges.     The  live  load  for 
railroad  bridges  varies  on  account  of  the  great  variations  in  spans 
and  wheel  spacings  of  engines.     The  bridges  for  main  tracks  are 
usually  designed  for  the  heaviest  engines  now  in  use,  or  that  may 
reasonably  be  expected  to  be  built  in  the  near  future. 


~1rt.ff<  LIVE    LOAD.  217 

The  live  load  may  be  treated  either  (a)  as  a  uniform  load,  (b) 
as  a  number  of  concentrated  wheel  loads  followed  by  a  uniform 
train  load,  or  (c)  as  an  equivalent  uniform  load.  The  second 
loading  is  preferable  and  gives  more  accurate  results,  but  requires 
more  work  to  determine  the  stresses. 

(a)  Uniform  Load.     When  train  loads  were  light,  it  was 
customary  to  design  the  bridges  for  a  uniform  load  instead  of 
considering  actual  wheel  concentrations ;    and  a  uniform  load  is 
still  often  used  in  practice.    The  same  load  is  generally  taken  for 
both  the  chord  and  the  web  numbers.     This  method  is  simpler 
than  that  involving  wheel  loads,  and  gives  fair  results  if  intelli- 
gently used.     Care  should  be  exercised,  however,  in  determining 
the  load  to  be  used  for  each  truss. 

(b)  Concentrated  Wheel  Loads.    The  method  of  using  con- 
centrated wheel  loads  is  complicated  by  the  great  variation  in  the 
weights  and  spacings  of  engine  wheel  loads.     Most  railroad  com- 
panies specify  that  the  stresses  shall  be  computed  for  two  engines 
and  tenders  followed  by  a  uniform  train  load. 

The  present  practice  among  many  railroad  companies  is  to  use 
a  conventional  wheel  loading,  one  of  the  best  examples  of  which 
being  that  given  by  Theodore  Cooper  in  his  "Specifications  for 
Steel  Railroad  Bridges  and  Viaducts".  The  three  common  classes 


o  oooo  oooo  o  oooo       oooo 

o  oooo  oooo  o  oooo 

o  oooo  lOioioLO  o  oooo 

in  oooo  <\j   CM     <NJ   <\J  10  oooo 

CM  10   m   m   in  rotoroto  oj  m   10   in   in 


o       oooo  m   to    in  in      o  oooo  in   in    in  LO  r.-^..  ., 

oooo  (vj   CM     <vj   r\j       i/>  oooo  cJcjcxjcJ  5000  IDS. 

10   m   m    in  roK>K~)rO       <M  m   u*>   in   in  ro  K~>     ro  ro  *»    r     ft 

QQQQ  QQQQ^Q  QQQQ 


L«i^£i*i-£i?^ 


Live   Load  per  Track  for  Cooper's  E>  50  Loading 
FIG.  116.     COOPER'S  E50  LOADING. 

recommended  by  Theodore  Cooper  are  his  £50,  £40,  and  £30. 
Fig.  116  shows  the  wheel  loads  and  spacings  for  Cooper's  £50 
loading,  which  corresponds  to  the  heaviest  engines  in  common 
use,  although  a  live  load  corresponding  to  an  E6o  loading  has 
been  used. 

An  £40  loading  has  the  same  wheel  spacings,  the  weight  of 


218 


LOADS. 


Chap.  XX. 


each  wheel  being  four-fifths  of  that  for  the  £50  loading,  followed 
by  a  uniform  train  load  of  4000  pounds  per  linear  foot  of  track. 

An  £30  loading  has  the  same  wheel  spacings,  the  weight  of 
each  wheel  being  three-fifths  of  that  for  the  £50,  followed  by  a 
uniform  train  load  of  3000  pounds  per  linear  foot  of  track. 


EQUIVALENT  UNIFORM  LOADS  -  COOPERS,  EL-4Q. 

5pan 

(Ft.) 

Equivalent  Uniform  Load 

Span 

(Ft-) 

Equivalent  Uniform  Load 

Chords 

Webs 

FI-Bms. 

Chords 

Webs 

Fl-Bms. 

10 

9000 

12000 

8200 

46 

6330 

7240 

5240 

11 

9310 

11640 

7960 

48 

6220 

7140 

5200 

12 

9340 

11330 

7830 

50 

61  10 

7060 

5140 

13 

9340 

11080 

7600 

52 

6040 

6940 

5130 

14 

9210 

10860 

7460 

54 

5960 

6820 

5IZO 

15 

9030 

10670 

7330 

56 

5880 

6720 

5110 

16 

8850 

10500 

7120 

58 

5800 

6620 

5090 

17 

6650 

10350 

6940 

60 

5730 

6530 

5080 

18 

8430 

10240 

6780 

62 

5690 

6490 

5080 

19 

8220 

10100 

6630 

64 

5700 

6450 

5070 

20 

8000 

10000 

6500 

66 

56ZO 

6450 

5070 

21 

8040 

9780 

6390 

68 

5560 

6380 

5060 

22 

8040 

9580 

6290 

70 

5510 

6340 

5060 

23 

8010 

9400 

6ZOO 

72 

5490 

6320 

5030 

24 

7960 

9230 

6120 

74 

5460 

6300 

5010 

25 

7890 

9080 

6040 

76 

5440 

6290 

4990 

26 

7780 

8930 

5970 

78 

5420 

6Z70 

4970 

27 

7660 

8790 

5900 

80 

5400 

6250 

4950 

28 

7540 

8660 

5B30 

82 

5370 

6230 

4930 

29 

7420 

8540 

5770 

84 

5340 

6200 

4910 

30 

7300 

8430 

5720 

86 

5310 

6180 

4890 

31 

7220 

8320 

5680 

86 

5270 

6150 

4870 

32 

7140 

8190 

5650 

90 

5250 

6130 

4860 

33 

7050 

8080 

5620 

92 

5250 

61  10 

4850 

34 

6960 

7980 

5600 

94 

5210 

6090 

4810 

35 

6870 

7890 

5570 

96 

5170 

6060 

4780 

36 

6820 

78ZO 

5530 

98 

5150 

6040 

4760 

37 

6760 

7750 

5500 

100 

5140 

60ZO 

4740 

38 

6700 

7690 

5460 

125 

5100 

5770 

4720 

39 

6630 

7630 

5430 

150 

5010 

5570 

4700 

40 

6560 

7570 

5400 

175 

4890 

5350 

4680 

42 

6530 

7450 

5340 

200 

4740 

5240 

4660 

44 

6470 

7340 

5300 

250 

4510 

5030 

4640 

FIG.  117.     EQUIVALENT  UNIFORM  LOADS. 

The  great  advantage  of  Cooper's  loadings  is  that  the  spacing 
of  wheels  is  the  same  for  all  loadings.     It  is  thus  seen  that  the 


Art.  3.  WIND   LOAD.  219 

moments  and  shears  for  the  different  loadings  are  proportional 
to  the  class  of  loading. 

The  actual  determination  of  moments,  shears,  and  stresses 
due  to  concentrated  wheel  loads  will  be  given  in  Chapter  XXIII. 

(c)  Equivalent  Uniform  Load.  An  equivalent  uniform  load 
is  one  which  will  give  results  closely  approximating  those  for 
actual  wheel  concentrations.  To  secure  accuracy,  it  is  necessary 
to  use  a  different  uniform  load  for  each  span  ;  also  a  different  load 
for  the  chords,  the  webs,  and  the  floorbeams.  The  table  in  Fig. 
117  gives  the  equivalent  uniform  loads  for  Cooper's  £40  loading. 
For  E5O  loading  use  125  per  cent  of  these  values,  and  for  E3O 
use  75  per  cent. 

The  stresses  in  trusses  due  to  equivalent  uniform  loads  may  be 
determined  by  the  methods  given  in  Chapter  XXI. 


ART.  3.    WIND  LOADS. 

The  wind  load  is  usually  expressed  in  one  of  the  two  follow- 
ing ways:  (a)  in  pounds  per  square  foot  of  actual  truss  surface; 
or  (b)   in  pounds  per  linear  foot,  treated  as  a  dead  load  acting 
upon  the  upper  and  lower  chords  and  as  a  live  load  acting  upon 
the  traffic  as  it  moves  over  the  bridge. 

1 68.  Wind  Load,     (a)     The  method  of  stating  the  wind 
load  in  pounds  per  square  foot  of  actual  truss  surface  has  the 
disadvantage  that  an  assumption  must  be  made  as  to  the  area  of 
the  truss  surface.    If  this  method  is  used,  the  wind  load  is  usually 
taken  at  30  pounds  per  square  fcot  of  truss  surface.     The  load 
may  be  considered  to  act  against  the  windward  truss  only,  or  to 
be  equally  resisted  by  the  two  trusses. 

169.  (b)     The   method   of   specifying  the   wind   load    in 
pounds  per  linear  foot  is  more  logical,  and  is  usually  employed. 

For  highway  bridges,  the  usual  practice  is  to  take  a  wind  load 
of  150  pounds  per  linear  foot,  treated  as  a  dead  load  acting  upon 
both  the  upper  and  lower  chords;  and  a  load  of  150  pounds  per 
linear  foot,  treated  as  a  live  load  acting  upon  the  portion  of  the 
bridge  covered  by  the  traffic. 


-  220  LOADS.  Chap.  XX. 

For  railroad  bridges,  a  higher  value  is  used  for  that  portion 
of  the  wind  which  is  treated  as  a  live  load.  This  is  to  provide, 
not  only  for  the  wind  loads,  but  also  for  stresses  caused  by  the 
vibrations  due  to  trains.  The  following-  wind  loads  are  recom- 
mended as  conforming  to  good  practice:  150  pounds  per  linear 
foot,  acting  upon  both  the  upper  and  lower  chords ;  and  450 
pounds  per  linear  foot  of  live  load  upon  the  bridge,  the  latter 
force  being  assumed  to  act  six  feet  above  the  base  of  the  rail. 
The  portion  of  the  wind  load  considered  as  live  load  will  be  re- 
sisted by  the  bottom  laterals  in  through  bridges,  and  by  the  top 
laterals  in  deck  bridges, 


CHAPTER  XXI. 
STEESSES  IN  TEUSSES  DUE  TO  UNIFOEM  LOADS. 

In  this  chapter  there  will  be  given  several  graphic  methods  for 
determining  the  stresses  in  various  types  of  bridge  trusses  under 
uniform  loads,  together  with  the  algebraic  method  of  coefficients. 
These  methods  will  be  explained  by  the  solution  of  particular 
problems ;  as  they  may  be  more  easily  understood  than  general 
ones.  Most  of  the  methods  are  of  general  application  to  the 
various  types  of  bridge  trusses;  but  the  problem  should  first  be 
studied  to  determine  which  method  may  be  employed  to  the  best 
advantage. 

The  chapter  will  be  divided  in  seven  articles,  as  follows :  Art. 
i,  Stresses  in  a  Warren  Truss  by  Graphic  Resolution';  Art.  2, 
Stresses  in  a  Pratt  Truss  by  Graphic  Resolution ;  Art.  3,  Stresses 
by  Graphic  Moments  and  Shears ;  Art.  4,  Stresses  in  a  Bowstring 
Truss — Triangular  Web  Bracing ;  Art.  5,  Stresses  in  a  Parabolic 
Bowstring  Truss ;  Art.  6,  Wind  Load  Stresses  in  Lateral  Sys- 
tems ;  and  Art.  7,  Stresses  in  Trusses  with  Parallel  Chords  by  the 
Method  of  Coefficients. 


ART.  i.    STRESSES  IN  A  WARREN  TRUSS  BY  GRAPHIC  RESOLUTION. 

The  application  of  the  method  of  graphic  resolution  to  the 
solution  of  the  stresses  in  a  WTarren  truss  will  now  be  explained. 

170.  Problem.  It  is  required  to  find  the  maximum  and 
minimum  dead  and  live  load  stresses  in  all  the  members  of  the 
highway  Warren  truss  shown  in  Fig.  118,  a.  The  truss  has  a 
span  L  of  70  feet ;  a  panel  length  1  of  10  feet ;  and  a  depth  D  of 
10  feet.  The  bridge  has  a  width  of  14  feet. 

221 


222 


STRESSES    IX    BRIDGE    TRUSSES. 


Chap.  XXI. 


171.  Dead  Load  Stresses.  The  dead  load  may  be  obtained 
by  applying  equation  (i),  §  163.  This  load  is  found  to  be  476  Ibs. 
per  ft.  of  span,  or  say  240  Ibs.  per  ft.  of  truss.  The  panel  load 
W  =  24oX  10  =  2400,  all  of  which  will  be  assumed  to  act  at 
the  lower  chord.  The  effective  reaction  ^=3X2400=7200. 

X 

/i  \Jz  Us  L/4  Us  Uz  u! 


IR, 

4 

L.             Lz             l_3    Y      ^ 

2                             X6,7     4 

(a)    L!2             LJ,              |Rz 

1                   »x 

A                   / 

x\  r 

x   \Z_ 

\  A     /:: 

6     A 

/  \                 / 

v  \ 

Y      5 
Y 

Y 

Y         0* 

\/  \  /  : 

5 

v      \  / 

3       (b)        \     / 

3000  *  400.0  *     i\Z_J  Y 

3          5       ^_] 

3  load  on  upper  chord  ,  |  on  lower.       Load  al  I  on  lower  chord  . 
D-IOft,  I  =10  ft-,  L=70  fT.jw  =  240lbs.perft. 

FIG.  118.     DEAD  LOAD  STRESSES  —  WARREN  TRUSS. 


To  determine  the  dead  load  stress,  lay  off  XY  =  72OO  (Fig,  118, 
b),  and  draw  the  dead  load  stress  diagram  for  one-half  of  the 
truss.  Since  the  truss  and  loads  are  symmetrical  and  the  full  dead 
load  is  always  on  the  bridge,  it  is  necessary  to  draw  the  diagram 
for  only  one-half  of  the  truss.  The  kind  of  stress  is  determined 
by  placing  the  arrows  on  the  truss  diagram  as  the  stress  diagram 
is  drawn.  The  dead  load  stresses  for  both  the  chord  and  web 
members  are  shown  in  Fig.  119,  a  and  Fig.  119,  b.  By  a  com- 
parison of  these  stresses,  it  is  seen  that,  for  this  truss,  the  two  web 
members  meeting  upon  the  unloaded  chord  have  the  same  nu- 
merical stress.  It  is  also  seen  that,  for  an  odd-panel  truss,  the 
center  web  members  are  not  stressed. 

The  assumption  is  sometimes  made  that  one-third  of  the  dead 
load  acts  at  the  upper  chord  and  two-thirds  at  the  lower  chord. 


Art.  1. 


WARREN     TRUSS GRAPHIC     RESOLUTION. 


223 


The  dead  load  stress  diagram  for  this  assumption  is  shown  in 
Fig.  1 1 8,  c.  For  highway  bridges,  it  is  customary  to  assume  all 
the  dead  load  on  the  lower  chord. 


MAXIMUM  AND  MINIMUM  STRESSES  IN  CHORD  MEMBERS 

Upper  Chord 

Lower   Chord 

Chord  Member 

X-2 

X-4 

X-6 

Y-l 

Y-3 

Y-5 

Y-7 

Dead  Load 
Live  Load 
Maximum 
Minimum 

-  7200 
-ZIOOO 
-Z8200 
-  7  ZOO 

-IZOOO 
-35000 
-47000 
-IZOOO 

-14400 
-4ZOOO 
-56400 
-14400 

+  3600 
+10500 
+14100 
+  3600 

+  9600 
+28000 
+37600 
+  9600 

+I3ZOO 
+38500 
+51700 
+I3ZOO 

+14400 
+4ZOOO 
+56400 
+14400 

(a) 


DEAD  LOAD  STRESSES  IN  WEB  MEMBERS 

Web  Member 

x-i 

hZ 

2-3 

3-4 

4-5 

5-6 

6-7 

Dead  Load 

-  8060 

+  8060 

-5370 

+  5370 

-2690 

+  Z690 

0 

(b) 


LIVE  LOAD  STRESSES  IN  WEB  MEMBERS 

Web  Member 

X-! 

1-2 

2-3 

3-4 

4-5 

5-6 

6-7 

Live  Load  at  Lli 

-    1120 

+  1120 

-  1120 

+  1120 

-  1120  ' 

+  HZO 

-1120 

..          „    Ll2 

-  2Z40 

+  2240 

-  2240 

+  2240 

-2Z40 

+  ZZ40 

-2Z40 

„       „  L!3 

-  3360 

+  3360 

-3360 

+  3360 

-3360 

+  3360 

-3360 

»       „  Ls 

-4480 

+  4480 

-4480 

+  4480 

-4480 

+  4480 

+  3360 

•»         «       "  Lz 

-5600 

+  5600 

-5600 

+  5600 

+  ZZ40 

-2Z40 

+  ZZ40 

n               ,,             „    L, 

-67ZO 

+  6720 

+  1120 

-  1120 

+  1120 

-IIZO 

+  IIZO 

Max-  Live  Load 

-235ZO 

+23520 

-16800 

+16800 

-IIZOO 

+  11  ZOO 

-67ZO 

Min-       "      » 

0 

0 

+  1120 

-  1  120 

+  3360 

-3360 

+  61ZO 

Uniform  »      " 

-Z35ZO 

+235ZO 

-15660 

+15680 

-  7840 

+  7840 

0 

(0 


MAXIMUM  AND  MINIMUM  STRESSES  IN  WEB  MEMBERS 

Web  Member 

X-I 

1-2 

2-3 

3-4 

4-5 

5-6 

6-7 

Dead  Load 
Max-  Live  Load 
Min-  Live  Load 
Max-  Stress 
Min-  Stress 

-  8060 
-Z35ZO 
0 
-31580 
-8060 

+  8060 
+235ZO 
0 
+31580 
+  8060 

-  5370 
-16800 
+  1120 
-ZZI70 
-4250 

+  5370 
+16800 
-  IIZO 
-ZZI70 
+  4Z50 

-  2690 
-11200 
+  3360 
-13890 
+    610 

+  2690 
+  11200 
-3360 
+13890 
-   670 

0 
-6720 
+  67ZO 
-6720 
+6720 

(d) 
Fia.  119.     TABLE  OP  STRESSES — WARREN  TRUSS. 

172.  Live  Load  Stresses.  The  live  load  will  be  taken  at 
1400  Ibs.  per  linear  ft.  of  bridge,  or  700  Ibs.  per  ft.  per  truss. 

Since  the  upper  and  lower  chords  are  parallel,  the  chords  will 
resist  the  bending  moment  due  to  the  loads,  and  the  web  members 
will  resist  the  shear. 


224 


STRESSES    IN    BRIDGE    TRUSSES. 


Chap.  XXI. 


(a)  Chord  Stresses.  It  is  seen  that  each  increment  of  live 
load  brought  upon  the  bridge  increases  the  bending  moment  and 
therefore  increases  the  stresses  in  each  chord  member.  The  maxi- 
mum live  load  stress  in  each  member  of  the  upper  and  lower 
chord  is  therefore  given  when  the  bridge  is  fully  loaded  with 
the  live  load.  The  minimum  stress  occurs  when  there  is  no  live 
load  on  the  bridge. 

Since  the  maximum  live  load  chord  stresses  are  obtained  when 
the  bridge  is  fully  loaded,  it  is  seen  that  it  is  unnecessary  to  draw 
a  new  stress  diagram ;  as  the  live  load  chord  stresses  may  be 
obtained  from  the  corresponding  dead  load  stresses  by  multi- 
plying each  stress  by  the  ratio  of  the  live  load  to  the  dead  load. 

The  maximum  live  load  chord  stresses  are  shown  in  Fig. 
119,  a.  The  minimum  live  load  chord  stresses  are  zero. 

X 


vww\x 


y    D=  10  ft-,  1=1  Oft-,  L=  10ft- 
p=  TOO  lbs-perft-iP=7000lb5. 

Stress  Diagram 

for 

Live  Load  at  LL«  only 
Ib) 


3'  5'   7   5   3   1 

FIG.  120.  LIVE  LOAD  STRESSES — WARREN  TRUSS. 

(b)  Web  Stresses.  By  partially  loading  the  truss  with  live 
load,  it  is  possible  to  obtain  stresses  which  are  larger  than  those 
obtained  when  the  bridge  is  fully  loaded ;  or  it  is  even  possible  to 
get  stresses  of  an  opposite  kind  to  those  due  to  the  dead  load. 


Art.l.  WARREN     TRUSS GRAPHIC     RESOLUTION.  225 

The  maximum  and  minimum  live  load  web  stresses  will  now  be 
found  by  applying  one  load  at  a  time,  and  noting  the  effect  of 
that  load  upon  the  stress  on  each  web  member. 

Fig.  1 20,  b  shows  the  stress  diagram  for  a  single  live  panel 

p 
load  placed  at  L/,  the  reaction  at  L0  being  — .     Since  the  truss 

is  symmetrical  about  the  center  line,  a  load  placed  at  L/  will  pro- 
duce the  same  stresses  in  the  web  members  to  the  left  of  the  cen- 
ter as  a  load  at  La  will  in  the  web  members  to  the  right  of  the 
center.  It  will  therefore  be  necessary  to  record  only  one-half  of 
the  web  members,  thus  simplifying  the  table  of  stresses.  From  the 
stress  diagram  (Fig.  120,  b),  it  is  seen  that  the  stresses  in  all  of 
the  members  to  the  left  of  L/  due  to  a  load  at  L/  are  constant. 

The  stresses  in  the  web  members  to  the  left  of  the  center  line 
of  the  truss  due  to  a  live  load  at  L/  are  shown  in  the  first  line 
of  Fig.  119,  c,  these  stresses  being  alternately  tension  and  com- 
pression. Now  place  a  live  load  at  L2' ',  the  remainder  of  the  truss 

being  unloaded.     The  reaction  at  L0  equals  —  P,  and  it  is  seen 

without  drawing  another  stress  diagram  that  the  stresses  in  the 
web  members  to  the  right  of  the  center  line  are  twice  those  due  to 
the  live  load  at  L/.  The  stresses  due  to  a  load  at  L2'  are  shown 
in  the  second  line  of  Fig.  119,  c.  It  is  also  seen  that  the  stresses 
for  a  load  at  L/  are  three  times  those  for  a  load  at  L/.  The 
stresses  for  a  load  at  L3'  are  shown  in  the  third  line  of  Fig.  119,  c. 
The  stresses  for  a  live  load  at  L3  are  shown  in  the  fourth  line ;  for 
a  live  load  at  L2  in  the  fifth  line ;  and  for  a  load  at  Lj  in  the  sixth 
line. 

Maximum  and  Minimum  Live  Load  Web  Stresses.  By  the 
maximum  live  load  stress  is  meant  the  greatest  live  load  stress 
that  ever  occurs  in  the  member;  and  by  the  minimum  live  load 
stress  is  meant  the  smallest  live  load  stress  if  there  is  no  reversal 
of  stress,  or  the  greatest  stress  of  an  opposite  kind  if  there  is  a 
reversal  of  stress  in  the  member. 

From  Fig.  119,  c,  it  is  seen  that  the  addition  of  each  load  pro- 
duces a  compressive  stress  in  X-i  and  a  tensile  stress  in  1-2,  the 
maximum  stress  being  —  23  520  for  X-i  and  +  23  520  for  1-2 


226  STRESSES    IK    BRIDGE    TRUSSES.  Chap.  XXL 

when  the  truss  is  fully  loaded.     The  minimum  live  load  stresses 
in  X-i  and  1-2  are  zero  when  there  is  no  live  load  on  the  truss. 

It  is  also  seen  that  loads  at  L/,  L2',  L3',  L3,  and  L2  all  produce 
compression  in  2-3  and  tension  in  3-4;  and  that  the  load  at  Lx 
produces  tension  in  2-3  and  compression  in  3-4.  The  maximum 
live  load  stresses  in  2-3  and  3-4  are  therefore  obtained  by  adding 
the  stresses  due  to  the  loads  at  L/,  L/,  L3',  L3,  and  L2,  and  are 
—  16800  and  +16800,  respectively,  (see  Fig.  119,  c).  The 
minimum  live-  load  stresses  in  2-3  and  3-4  are  obtained  when 
there  is  a  live  load  at  L,.  only,  and  are  +  1120  and  — 1120, 
respectively. 

The  maximum  live  load  stresses  in  4-5  and  5-6  are  obtained 
when  the  loads  at  L/,  L/,  L3',  and  L3  are  on  the  bridge,  and  are 
• —  ii  200  and  +11  200,  respectively.  The  minimum  live  load 
stresses  in  4-5  and  5-6  are  obtained  when  the  loads  at  L±  and  L2 
are  on  the  bridge,  and  are  +  3360  and  —  3360,  respectively. 

The  maximum  live  load  stress  in  6-7  is  obtained  when  the 
loads  at  L/,  L/,  and  L/  are  on  the  bridge,  and  is  —  6720.  The 
minimum  live  load  stress  in  6-7  is  obtained  when  the  loads  at 
Lj,  L.,,  and  L3  are  on  the  bridge,  and  is  +  6720. 

A  comparison  of  the  corresponding  stresses  in  line  7  and  line 
9  shows  the  difference  in  the  stresses  in  each  member  for  the 
bridge  loaded  for  maximum  live  load  stresses,  and  fully  loaded 
with  live  load. 

173.  Maximum  and  Minimum  Dead  and  Live  Load 
Stresses,  (a)  Chord  Stresses.  The  maximum  and  minimum 
chord  stresses  due  to  dead  and  live  loads  are  shown  in  Fig.  119,  a. 
The  maximum  chord  stresses  are  obtained  when  the  bridge  is 
fully  loaded  with  dead  and  live  loads,  and  are  equal  to  the  sum  of 
the  dead  and  live  load  stresses.  The  minimum  chord  stresses  are 
obtained  when  there  is  no  live  load  on  the  bridge,  and  are  the  dead 
load  stresses. 

(b)  Web  Stresses.  The  maximum  and  minimum  web  stresses 
due  to  dead  and  live  loads  are  shown  in  Fig.  119,  d.  The  maxi- 
mum web  stresses  are  obtained  by  adding  the  dead  and. maximum 
live  load  stresses ;  since  they  have  the  same  signs. 


Art.l.  WARREN    TRUSS — GRAPHIC    RESOLUTION.  227 

The  minimum  web  stresses  are  obtained  by  adding,  alge- 
braically, the  dead  and  minimum  live  load  stresses.  It  is  seen 
that  the  dead  and  minimum  live  load  stresses  have  opposite  signs. 

By  comparing  the  dead  and  the  minimum  web  stresses,  it  is 
seen  that  the  members  4-5,  5-6,  and  6-7  have  reversals  of  stress. 

174.  Loadings    for    Maximum    and    Minimum    Stresses. 
From  a  comparison  of  the  stresses  in  Fig.  119,  the  following  con- 
clusions may  be  drawn  for  the  maximum  and  minimum  dead  and 
live  load  stresses  in  a  Warren  truss. 

(a)  For  maximum  chord  stresses,  load  the  bridge  fully  with 
dead  and  live  loads. 

(b)  For  minimum  chord  stresses,  load  the  bridge  with  dead 
load  only. 

(c)  For  maximum  web  stress  in  any  member,  load  the  longer 
segment  of  the  bridge  with  live  load,  the  shorter  segment  being 
unloaded. 

(d)  For   minimum  zveb   stress   in   any   member,   load    the 
shorter  segment  of  the  bridge  with  live  load,  the  longer  segment 
being  unloaded. 

175.  Simplified  Construction  for  Live  Load  Web  Stresses. 
Referring  to  Fig.  120,  b,  it  is  seen  that  the  numerical  stresses  in 
all  the  web  members  to  the  left  of  any  lower  chord  panel  point 
are  constant  when  the  live  load  extends  to  the  right  of  that  panel 
point.     As  the  stresses  are  alternately  tension  and  compression, 
it  is  seen  that  it  is  only  necessary  to  draw  the  triangle  YaY  (Fig. 
1 20,  b).     This  triangle  is  constructed  by  making  YY  equal  to 
one  live  panel  load,*  and  drawing  Ya  parallel  to  be  to  meet  a 
horizontal  line  Ya.    The  inclination  of  be  is  determined  by  mak- 
ing Yb  equal  to  (or  some  multiple  of)  the  half  panel  length,  and 
Yc  equal  to  (or  the  same  multiple  of)  the  depth  of  the  truss.    The 
stresses  in  the  members  to  the  right  of  L/  when  there  is  a  load 

at  L/  are  equal  to  —  Ya.  When  there  is  a  load  at  L2',  the  stresses 
to  the  right  of  L/  are  equal  to  —  Ya,  etc.  It  is  therefore  unneces- 
sary to  draw  the  entire  diagram  shown  in  Fig.  120,  b. 


228 


STRESSES    IX    BRIDGE    TRUSSES. 


Chap.  XXI. 


ART.  2.    STRESSES  IN  A  PRATT  TRUSS  BY  GRAPHIC  RESOLUTION. 

176.  The  stresses  in  a  through  Pratt  truss  will  be  deter- 
mined in  this  article.  It  was  shown  in  the  preceding  article  that 
some  of  the  web  members  of  the  Warren  truss  were  subjected  to 
reversals  of  stress.  In  the  Pratt  truss  shown  in  Fig.  121,  a,  the 
web  members  are  so  constructed  that  they  can  not  resist  reversals 
of  stress,  the  intermediate  posts  taking  compression  only,  and  the 
intermediate  diagonals,  tension  only.  This  involves  the  use  of 
another  set  of  diagonals,  called  counters,  in  those  panels  where 
there  is  a  tendency  for  reversals. 


U 


5  load  on  upper  chord,  |  on  lower.      Load  all  on  lower  chord. 


D  =  28ft.,  l-ZOft-j  L-  140  ft-;  w=7IO  Ibs-perft- 

FIG.  121.     DEAD  LOAD  STRESSES  —  P.UATT  TRUSS. 

The  two  center  diagonals  of  the  seven-panel  truss  shown  in 
Fig.  121,  a  are  both  counters;  as  there  is  no  dead  load  shear  in 
the  center  panel  of  an  odd-panel  truss.  The  counter  U3L3'  acts 
when  the  live  load  extends  to  the  right  of  L3',  and  the  counter 
U3'L3,  when  the  live  load  extends  to  the  left  of  L3. 

Care  must  be  taken  in  determining  the  stresses  in  the  verticals 
adjacent  to  the  counters  ;  as  they  differ  greatly  from  those  which 
would  exist  in  these  members  if  the  main  ties  could  resist  com- 
pression. 


PRATT  TRUSS GRAPHIC  RESOLUTION. 


229 


177.     Problem.     It  is  required  to  find  the  maximum  and 
minimum  dead  and  live  load  stresses  in  the  through  Pratt  truss 


MAXIMUM  AND  MINIMUM  STRESSES  IN  CHORD  MEMBERS  - 

Upper  Chord 

Lower  Chord 

Chord  Member 

X-3 

X-5 

X-6 

Y-l 

Y-2 

Y-4 

Y-6 

Dead  Load 
Live  Load 
Maximum 
Minimum 

-  50.7 
-182.8 
-233.5 
-  50.7 

-  60.8 
-219.2 
-280.0 
-  60.8 

-  60.8 
-219.2 
-Z80.0 
-  60.8 

+  30.4 
+109.6 

+  140.0 
+  30.4 

+  30.4 
+  109.6 
+  140.0 
+  30.4 

+  50.7 
+182.8 
+233.5 
+  50.7 

+  60.8 
+  219.2 
+  280.0 
+  60.8 

(a) 


DEAD  LOAD  STRESSES  IN  WEB  MEMBERS 

Web  Member 

x-i 

1-2 

2-3 

3-4 

4-5 

5-6 

6-6' 

Dead  Load 

-52.4 

+  14.2 

+  34.9 

-14.2 

+  17.5 

0.0 

0.0 

(b) 


LIVE 

LOAD  ' 

STRESS 

>ES  IN 

WEB  t 

^IEMBE 

R5 

Web  Member 

X-l 

1-2 

2-3 

3-4 

4-5 

5-6 

6-6' 

Live  Load  at  lli 

-    9.0 

0.0 

+    9.0 

-    7.3 

+    9.0 

-   7.3 

+    9.0 

»  L'z 

-  18.0 

0.0 

+  18.0 

-  14.6 

+  18.0 

-14.6 

+  18.0 

*  & 

-  27.0 

0.0 

+  27.0 

-21.9 

+  27.0 

-21.9 

+  Z7.0 

»  L* 

-36.0 

0.0 

+  36.0 

-  29.2 

+  36.0 

+  21.9 

-  27.0 

••  Lz 

-45.0 

0.0 

+  45.0 

+  14.6 

-  18.0 

+  14.6 

-  18.0 

»»                ••  Li 

-  54.0 

+  51.2 

-    9.0 

+    7.3 

-    9.0 

+  7.3 

-  9.0 

Max-  Live  Load 

-189.0 

+  51.2 

+135.0 

-73.0 

+  90.0 

-43.b 

+  54.0 

Min. 

0.0 

0.0 

-    9.0 

+  21.9 

-27.0 

+43.8 

-54.0 

Uniform  »»       » 

-189.0 

+  51.2 

+126.0 

-  51.2 

+  63.0 

0.0 

0.0 

(C) 


MAXIMUM  AND  MINIMUM  STRESSES  IN  WEB  MEMBERS 

Web  Member 

X-l 

1-2 

2-3 

3-4 

4-5 

4-5 

Counter 

5-6 

6-6' 

Counter 

Dead  Load 
Max-  Live  Load 
Min-  Live  Load 
Max-  Stress 
Min-  Stress 

-  524 
-189.0 
0.0 
-241.4 
-52.4 

+  14.2 
+  51.2 
0.0 
+  65.4 
+  I4.Z 

+  34.9 
+  135.0 
-   9.0 
+169.9 
+  25.9 

-  142 
-73.0 
+  21.9 
-87.2 

o.o 

+  17.5 
+90.0 
-27.0 
+107.5 
0.0 

-1-  9.5 
0.0 

0.0 
-43.8 
+  43.8 
-43.8 
0.0 

0.0 
+  54.0 
-54.0 
+  54.0 
0.0 

All  stresses  are  in  thousands  of  pounds 

(d) 
FIG.  122.     TABLE  OF  STRESSES — PRATT  TRUSS. 

shown  in  Fig.  121,  a.    The  truss  has  a  span  of  140  feet,  a  panel 
length  of  20  feet,  and  a  depth  of  28  feet. 

178.     Dead  Load  Stresses.    The  dead  load  will  be  taken  at 
710  Ibs.  per  ft.  per  truss.  The  panel  load  W  =  710  X  20  =  14  200. 


230  STRESSES    IX    BRIDGE    TRUSSES.  Chap.  XXL 

The  effective  reaction  Rx  =  14200  X  3=42600.  In  this  prob- 
lem, all  the  dead  load  will  be  taken  on  the  lower  chord  of  the 
truss.  The  stresses  will  be  expressed  in  thousands  of  pounds, 
being  carried  to  the  nearest  100  pounds;  thus,  14200  will  be 
written  14.2. 

The  dead  load  stress  diagram  for  the  left  half  of  the  truss  is 
shown  in  Fig.  121,  b.  Since  the  truss  and  loads  are  symmetrical 
with  respect  to  the  center  line,  it  is  only  necessary  to  draw  the 
diagram  for  one-half  of  the  truss. 

The  dead  load  chord  stresses  are  shown  in  Fig.  122,  a,  and 
the  dead  load  web  stresses  in  Fig.  122,  b.  The  stresses  for  the 
left  half  of  the  truss  only  are  shown ;  as  those  for  the  right  half 
are  equal  to  the  -stresses  in  the  corresponding  members  shown. 
It  is  seen  that  the  upper  and  lower  chord  stresses  increase  from 
the  end  toward  the  center  of  the  truss,  and  that  the  web  stresses 
decrease  from  the  end  toward  the  center.  It  is  further  seen  that 
the  stresses  in  the  web  members  in  the  center  panel  of  this  truss 
are  zero,  and  that  the  stresses  in  the  three  center  panels  of  the 
upper  chord  are  equal. 

Fig.  121,  c  shows  the  dead  load  stress  diagram  for  the  left 
half  of  the  truss,  assuming  that  one-third  of  the  dead  load  is  on 
the  upper  chord  and  two-thirds  on  the  lower  chord.  By  compar- 
ing the  stress  diagrams  shown  in  Fig.  121,  b  and  Fig.  121,  c,  it  is 
seen  that  the  chord  and  inclined  web  stresses  are  the  same  for  both 
cases,  and  that  the  stresses  in  the  vertical  posts,  only,  are  changed. 
When  one-third  of  the  dead  load  is  taken  on  the  upper  chord, 
the  stresses  in  the  intermediate  posts  are  greater  by  the  amount 
of  the  upper  chord  panel  load  than  when  all  the  load  is  taken  on 
the  lower  chord;  while  the  stress  in  the  hip  vertical  is  smaller 
by  the  amount  of  the  upper  chord  panel  load. 

179.  Live  Load  Stresses.  The  live  load  will  be  taken  at 
2560  Ibs.  per  ft.  per  truss.  The  panel  load  P  —  2560  X  20  = 
51  200,  or  say  51.2. 

(a)  Chord  Stresses.  Since  the  addition  of  each  increment  of 
live  load  increases  the  bending  moment,  and  since  the  upper  and 
lower  chords  in  trusses  with  parallel  chords  must  resist  this  bend- 
ing moment,  it  is  seen  that  the  maximum  stresses  in  all  the  chord 


Art. 


PRATT    TRUSS — GRAPHIC    RESOLUTION". 


231 


members  are  obtained  when  the  bridge  is  fully  loaded  with  live 
load. 

The  minimum  live  load  chord  stresses  are  zero,  when  there  is 
no  live  load  on  the  bridge. 

The  chord  stresses  may  be  obtained  either  by  loading  the 
bridge  fully  with  live  load  and  drawing  a  stress  diagram  similar 
to  that  shown  in  Fig.  121,  b,  or  they  may  be  obtained  from  the 
dead  load  chord  stresses  by  direct  proportion.  Each  chord  stress 
is  equal  to  the  corresponding  dead  load  stress  multiplied  by  the 


u, 


u; 


Stress  Diagram 
for 

Live  Load  at  Li  only 
ib) 


4'  64      1,2 

FIG.  123.     LIVE  LOAD  STRESSES — PRATT  TRUSS. 

ratio  of  the  live  load  to  the  dead  load.  The  latter  method  requires 
less  work  and  was  used  in  this  problem.  The  maximum  live  load 
chord  stresses  are  shown  in  Fig.  122,  a. 


232  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

(b)  Web  Stresses.  The  web  stresses,  and  the  positions  of 
the  loads  for  maximum  and  minimum  live  load  web  stresses,  will 
now  be  found  by  applying  one  load  at  a  time  and  obtaining  the 
stress  in  each  member  due  to  that  load. 

The  stress  diagram  for  a  single  live  load  at  L/  is  shown  in 
Fig.  123,  b.  Since  there  are  seven  panels  in  this  truss,  the  re- 
action at  L0  will  be  equal  to—  P,  and  the  reaction  at  L0',  to 

—P.     This  diagram  is  constructed  by  laying  off  Rx  =  —  P,  and 

drawing  the  stress  diagram  for  the  entire  truss  (see  Fig.  123,  b). 
In  drawing  this  diagram,  it  is  assumed  that  the  diagonals  which 
are  stressed  by  the  dead  load,  and  the  diagonal  6-6'  (U3L3')  are 
acting.  This  set  of  diagonals  will  act  unless  the  dead  load  stresses 
in  some  of  the  members  are  reduced  to  zero  by  the  live  load,  thus 
throwing  some  of  the  counters  into  action. 

Since  the  stresses  in  $'-4',  5-6,  and  3-4  are  numerically  equal, 
as  are  also  those  in  2'~3',  4'-$',  5 '-6,  5-4,  3-2,  and  X-i,  for  a 
single  load  at  L/,  it  is  unnecessary  to  draw  the  entire  diagram 
shown  in  Fig.  123,  b.  The  stresses  in  these  members  may  be 
obtained  from  the  triangle  YaYj.  This  triangle  is  constructed  by 
laying  off  YYt  equal  to  the  live  panel  load  P,  and  drawing  Ya 
parallel  to  be  to  meet  a  horizontal  line  through  Yt.  The  inclina- 
tion of  be  is  determined  by  laying  off  Ytb  equal  to  some  multiple 
(in  this  case  the  multiple  is  2)  of  the  panel  length,  and  YjC  equal 
to  the  same  multiple  of  the  depth  of  the  truss.  For  a  single  live 

load  at  L/,  the  reaction  at  L0  is  equal  to  —  P,  and  the  stresses  in 

the  inclined  diagonals  are  each  equal  to  —  Ya;  while  those  in  the 
verticals,  with  the  exception  of  the  hip  vertical,  are  each  equal  to 
—  YY±.  The  stress  in  the  hip  vertical  i'-2'  is  not  influenced  by 

any  of  the  loads  on  the  truss  except  that  at  L/,  and  therefore  the 
stress  in  this  member  is  either  o  or  P. 

The  stresses  in  the  web  members  to  the  left  of  the  center  of  the 


Art.  2.         PRATT  TRUSS — GRAPHIC  RESOLUTION.  233 

truss  when  there  is  a  live  load  at  L/  are  shown  in  the  first  line 
of  Fig.  122,  c.  It  is  seen  that  the  stress  in  1-2  is  zero. 

When  there  is  a  single  live  load  at  L./,  the  reaction  at  L0  is 

sy 

equal  to  —  P,  and  the  stresses  in  the  inclined  and  vertical  web 

2  2 

members    (1-2  excepted)    are  each  equal  to  ^    Ya  and—  YYX, 

respectively.  The  stresses  in  the  web  members  to  the  left  of  the 
center  when  there  is  a  live  load  at  L/  are  shown  in  the  second 
line  of  Fig.  122,  c. 

When  there  is  a  single  live  load  at  L3',  the  reaction  at  L0  is 

equal  to  —  P,  and  the  stresses  in  the  web  members  to  the  left  of 

the  center  for  this  loading  are  shown  in  the  third  line  of  Fig. 
122,  c. 

The  stresses  in  the  web  members  to  the  left  of  the  center 
when  there  is  a  single  live  load  at  L3  are  shown  in  the  fourth 
line  of  Fig.  122,  c ;  when  there  is  a  single  live  load  at  L2,  in  the 
fifth  line ;  and  when  there  is  a  single  live  load  at  L15  in  the  sixth 
line. 

Maximum  and  Minimum  Live  Load  Web  Stresses.  By  com- 
paring the  web  stresses  shown  in  Fig.  122,  b  with  those  shown  in 
Fig.  122,  c,  it  is  seen  that  the  live  load  stresses  in  the  members  to 
the  left  of  the  center  due  to  loads  at  L/,  L/,  and  L3'  have  the 
same  signs  as  the  corresponding  dead  load  stresses.  It  is  also 
seen  that  loads  placed  at  L3,  L2,  and  L1?  successively,  produce  the 
same  kinds  of  stress  in  the  members  to  the  left  of  these  points 
as  does  the  dead  load. 

A  live  load  placed  at  Lt  tends  to  produce  an  opposite  kind  of 
stress  in  the  members  2-3,  3-4,  and  4-5  to  that  caused  by  the 
dead  load ;  and  a  load  placed  at  L2  tends  to  produce  an  opposite 
kind  of  stress  3-4  and  4-5. 

Since  in  this  truss  the  intermediate  diagonals  are  tension  mem- 
bers, the  loads  at  and  to  the  right  of  L./  cause  the  counter  U3L3' 
to  act ;  and  loads  at  and  to  the  left  of  L3  cause  the  counter  U3'L3 
to  act. 

From  Fig.  122,  c,  it  is  seen  that  the  addition  of  each  live  load 


234  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

produces  a  compressive  stress  in  X-i,  the  maximum  stress  being 

—  189.0,  which  is  the  sum  of  the  stresses  caused  by  the  separate 
loads.    The  minimum  live  load  stress  in  X-i  is  o,  when  there  is 
no  live  load  on  the  bridge. 

It  is  also  seen  from  Fig.  122,  c  that  the  maximum  live  load 
stress  in  the  hip  vertical  1-2  is  obtained  when  there  is  a  live 
load  at  L!,  and  is  equal  to  the  load  at  that  point.  This  member 
simply  transfers  the  load  to  the  joint  U15  and  the  stress  in  it  is 
not  influenced  by  any  other  loads  on  the  truss.  The  minimum  live 
load  stress  is  o,  when  there  is  no  live  load  on  the  bridge. 

The  maximum  live  load  stress  in  the  member  2-3  is  obtained 
when  the  loads  at  and  to  the  right  of  L2  are  on  the  truss,  and 
is  -f-  135.0,  the  sum  of  the  separate  stresses  caused  by  these  loads. 
The  minimum  live  load  stress  in  2-3  is  obtained  when  the  single 
live  load  at  Lx  is  on  the  truss,  and  is  — 9.0  (provided  there  is 
already  that  much  tension  in  the  member  due  to  the  dead  load). 

The  maximum  live  load  stress  in  the  vertical  3-4  is  obtained 
when  the  loads  at  and  to  the  right  of  L3  are  on  the  truss,  and  is 

—  73.0.     The  minimum  live  load  stress  in  3-4  is  obtained  when 
the  loads  at  Lj,  and  L2  are  on  the  truss,  and  is  +21.9  (provided 
the  counter  in  the  panel  L2  L3  does  not  act  for  this  loading). 

The  maximum  live  load  stress  in  4-5  is  obtained  when  the 
loads  at  and  to  the  right  of  L3  are  on  the  truss,  and  is  +  9°-°- 
The  minimum  live  load  stress  in  4-5  is  obtained  when  the  loads 
at  Lj.  and  L2  are  on  the  truss,  and  is  — 27.0  (provided  there  is 
that  much  dead  load  tension  already  in  it). 

The  maximum  live  load  stress  in  5-6  is  obtained  when  the 
loads  at  L/,  L2',  and  L3'  are  on  the  truss,  and  is  — 43.8.  The 
minimum  live  load  stress  in  5-6  is  obtained  when  the  loads  at 
Lx,  L2,  and  L3  are  on  the  truss,  also  when  the  truss  is  fully  loaded. 

The  maximum  live  load  stress  in  6-6'  (U3L3')  is  obtained 
when  the  loads  at  L/,  L/,  and  L3'  are  on  the  truss,  and  is  +  54-°- 
The  minimum  live  load  stress  in  6-6'  is  obtained  when  the  loads 
at  Lt,  L2  and  L3  are  on  the  truss,  also  when  the  truss  is  fwlly 
loaded. 

By  comparing  the  above  stresses,  the  following  conclusions 
may  be  drawn:  (a)  there  is  no  stress  in  the  hip  vertical  1-2 


Art.  2.         PRATT  TRUSS — GRAPHIC  RESOLUTION.  235 

unless  there  is  a  load  at  L1?  and  when  there  is  a  load  at  this  point, 
the  stress  in  the  hip  vertical  is  equal  to  that  load;  (b)  the  web 
members  meeting  on  the  unloaded  chord  have  their  maximum 
and  minimum  live  load  stresses  under  the  same  loading;  (c)  the 
maximum  live  load  web  stresses  are  obtained  when  the  longer 
segment  of  the  truss  is  loaded  with  live  load;  (d)  the  minimum 
live  load  web  stresses  are  obtained  when  the  shorter  segment  of 
the  truss  is  loaded. 

A  comparison  of  the  stresses  shown  in  line  7  and  line  9  (Fig. 
122,  c)  shows  the  differences  in  the  corresponding  stresses  for 
the  truss  loaded  for  maximum  live  load  stresses  and  fully  loaded 
with  live  load. 

1 80.  Maximum  and  Minimum  Dead  and  Live  Load 
Stresses,  (a)  Chord  Stresses.  The  maximum  and  minimum 
chord  stresses  due  to  dead  and  live  loads  are  shown  in 
Fig.  122,  a.  The  maximum  chord  stresses  are  obtained  when 
the  truss  is  fully  loaded  with  dead  and  live  loads,  and  are  equal 
to  the  sums  of  the  corresponding  dead  and  live  load  stresses.  The 
minimum  chord  stresses  are  obtained  when  there  is  no  live  load 
on  the  bridge,  and  are  the  dead  load  stresses. 

(b)  Web  Stresses.  The  maximum  and  minimum  web 
stresses  due  to  live  and  dead  loads  are  shown  in  Fig.  122,  d.  Since 
the  intermediate  posts  take  compression  only,  and  the  intermediate 
diagonals,  tension  only,  care  must  be  used  in  making  the  com- 
binations for  maximum  and  minimum  stresses.  It  should  be  borne 
in  mind  that  the  counter  and  main  tie  in  any  panel  cannot  act  at 
the  same  time,  and  that  the  counter  does  not  act  until  the  dead 
load  tension  in  the  main  tie  in  that  panel  has  been  reduced  to 
zero  by  the  live  load.  It  is  thus  seen  that  a  main  tie  can  resist 
as  much  live  load  compression  as  there  is  dead  load  tension 
already  in  it,  and  no  more,  the  remainder  of  the  live  load  stress,  if 
any,  being  taken  by  the  counter.  Since  the  dead  load  always  acts, 
it  must  be  considered  in  all  combinations. 

Referring  to  Fig.  122,  c  and  Fig.  122,  d,  it  is  seen  that  the 
maximum  stress  in  X-i  is  obtained  when  the  bridge  is  fully 
loaded  with  live  and  dead  loads,  and  is  — 241.4.  The  minimum 


236  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

stress  in  X-i  is  obtained  when  there  is  no  live  load  on  the  bridge, 
and  is  —  52.4. 

The  maximum  stress  in  the  hip  vertical  1-2  is  obtained  when 
there  is  a  live  load  at  L±,  and  is  -f-  65.4,  the  sum  of  the  dead  and 
live  panel  loads.  The  minimum  stress  in  1-2  is  obtained  when 
there  is  no  live  load  on  the  bridge,  and  is  +  14.2. 

The  maximum  stress  in  2-3  (see  Fig.  122,  c  and  Fig.  .122,  d) 
is  obtained  when  there  are  live  loads  at  and  to  the  right  of  L2, 
and  is  -j~  169-9-  The  minimum  stress  in  2-3  is  obtained  when 
there  is  a  live  load  at  Lx  only,  and  is  +  25.9.  In  this  case,  it 
is  seen  that  the  minimum  live  load  stress  has  an  opposite  sign 
from  that  of  the  dead  load,  and  subtracts  from  it.  No  counter 
is  required  in  this  panel,  as  the  live  load  compression  is  less  than 
the  dead  load  tension. 

The  stress  in  the  member  4-5  will  be  determined  before  that 
in  3-4,  as  it  is  necessary  to  determine  which  diagonal  is  acting 
before  the  stress  in  the  post  can  be  found. 

The  maximum  stress  in  4-5  is  obtained  when  the  loads  at 
and  to  the  right  of  L3  are  on  the  bridge,  and  is  -f-  107.5.  The 
minimum  stress  in  4-5  is  obtained  when  the  live  loads  at  L±  and 
L2  are  on  the  bridge,  and  is  o ;  as  will  now  be  shown.  Referring 
to  Fig.  122,  c,  it  is  seen  that  the  live  load  tends  to  cause  a  com- 
pression of  27.0  in  4-5  when  the  loads  at  L^  and  L2  are  on  the 
bridge.  The  member  can  only  resist  a  compression  of  17.5  (this 
being  the  dead  load  tension  already  in  it),  the  resulting  stress  in 
4-5  being  o.  The  remainder  of  the  live  load  stress  tends  to  dis- 
tort the  member  4-5,  and  throws  the'  counter  in  this  panel  into 
action.  The  stress  in  the  counter  is  then  +  9.5. 

The  maximum  stress  in  3-4  is  obtained  when  the  live  loads 
at  and  to  the  right  of  L3  are  on  the  bridge,  and  is  —  87.2.  The 
minimum  stress  in  3-4  is  obtained  when  the  live  loads  at  Lt 
and  L2  are  on  the  bridge,  and  is  o.  It  is  seen  that  the  stress  in 
3-4  must  be  o ;  as  the  stress  in  4-5  has  been  shown  to  be  o  for 
this  loading,  and  for  equilibrium  at  U2,  the  stress  in  3-4  must 
also  be  o. 

The  maximum  stress  in  the  counter  6-6'  (U3L3')  is  obtained 
when  there  are  live  loads  at  L3',  L/,  and  L/,  and  is  +  54-°- 


Art.  2.         PRATT  TRrSS GRAPHIC  RESOLUTION.  237 

The  minimum  stress  in  6-6'  is  obtained  when  there  is  no  live 
load  on  the  bridge,  when  the  bridge  is  fully  loaded,  or  when  the 
live  loads  at  Lx,  L2,  and  L3  are  on  the  span,  and  is  o.  There  are 
no  dead  load  stresses  in  the  center  diagonals  of  this  truss. 

The  maximum  stress  in  5-6  is  obtained  when  there  are  live 
loads  at  L3',  L/,  and  L/,  and  is  — '43-8.  The  minimum  stress 
in  5-6  is  obtained  when  there  is  no  live  load  on  the  bridge,  also 
when  the  live  loads  at  L1?  L2,  and  L3  are  on  the  bridge,  and  is  o. 

The  maximum  stress  in  the  other  counter  L3U3'  is  obtained 
when  the  loads  at  L1?  L2,  and  L3  are  on  the  bridge,  and  is  -j-  54.0. 
The  minimum  stress  is  obtained  when  there  are  no  live  loads  on 
the  bridge,  when  the  bridge  is  fully  loaded,  or  when  there  are 
live  loads  at  L/,  L/,  and  L/. 

It  is  unnecessary  to  determine  the  stresses  in  the  members 
beyond  the  center  line ;  as  the  truss  is  symmetrical. 

181.  Loadings    for   Maximum    and   Minimum    Stresses. 
Conclusions.     From  a  comparison  of  the  stresses  shown  in  Fig. 
122,  the  following  conclusions  may  be  drawn  for  maximum  and 
minimum  dead  and  live  load  stresses  in  a  Pratt  truss. 

(a)  For  maximum  chord  stresses,  load  the  bridge  fully  with 
live  and  dead  loads. 

(b)  For  minimum  chord  stresses,  load  the  bridge  with  dead 
lo'ad  only. 

(c)  For  maximum  web  stress  in  any  member  (except  the  hip 
vertical),  load  the  longer  segment  of  the  truss  with  live  load.  For 
maximum  stress  in  the  hip  vertical,  load  the  bridge  so  that  there 
will  be  a  live  load  at  L^. 

(d)  For  minimum  web  stress  in  any  member  (except  the  hip 
vertical),  load  the  shorter  segment  of  the  bridge  with  live  load. 
For  minimum  stress  in  the  hip  vertical,  load  the  bridge  ivith  dead 
load  only. 

In  making  the  combinations  for  maximum  and  minimum 
stresses,  it  should  be  borne  in  mind  that  the  counter  in  any  panel 
does  not  act  until  the  dead  load  stress  in  the  main  diagonal  in  that 
panel  has  been  reduced  to  zero  by  the  live  load. 

182.  The  stresses  in  the  members  of  a  Howe  truss  may  be 
determined  in  a  similar  manner  to  that  used  in  finding  those  in  a 


238 


STRESSES    IN    BRIDGE    TRUSSES. 


Chap.  XXL 


Pratt  truss.    In  the  Howe  truss,  the  vertical  members  take  tension 
only,  and  the  diagonals,  compression  only. 


ART.  3.    STRESSES  BY  GRAPHIC  MOMENTS  AND  SHEARS. 

The  method  of  graphic  moments  and  shears,  as  applied  to 
the  solution  of  the  stresses  in  a  bridge  truss,  will  be  explained  in 
this  article. 

183.  Problem.  It  is  required  to  find  the  stresses  in  the 
six-panel  Warren  truss  shown  in  Fig.  124,  a.  The  truss  has  a 
span  of  1 20  feet,  a  panel  length  of  20  feet,  and  a  depth  of  20  feet. 
The  dead  load  will  be  taken  at  800  Ibs.  per  ft.  per  truss,  and  the 
live  load  at  1600  Ibs.  per  ft.  per  truss. 

X 


.i  W=  16000  Ibs. 
FIG.  124.     DEAD  LOAD  STRESSES  BY  GRAPHIC  MOMENTS  AND  SHEARS. 

184.     Dead  Load  Chord  Stresses.    Draw  the  truss  diagram 
(Fig.  124,  a)  to  scale,  and  load  it  fully  with  dead  load.    The  dead 


Art.  3.  GRAPHIC    MOMENTS    AND    SHEARS.  239 

panel  load  is  equal  to  800X20=16000.  Draw  the  load  line 
shown  in  Fig.  124,  c  for  the  dead  load.  Take  the  pole  O  with 
a  pole  distance  H,  draw  the  rays,  and  construct  the  funicular 
polygon  shown  in  Fig.  124,  b.  The  pole  distance  H,  expressed 
in  thousands  of  pounds,  should  be  numerically  equal  to  the  depth 
of  the  truss,  or  to  some  multiple  of  the  depth,  i.  e.,  if  D  =  20, 
then  H  should  be  equal  to  20  ooo,  or  to  some  multiple  of  20  ooo. 
The  vertices  of  the  funicular  polygon  shown  in  Fig.  124,  b  lay  on 
the  arc  of  the  bending  moment  polygon  for  the  given  truss  and 
loads.  The  bending  moment  in  any  upper  chord  member  is  equal 
to  the  intercept  under  the  center  of  moments  multiplied  by  the 
pole  distance  H ;  and  the  stress  in  the  member  is  equal  to  the 
bending  moment  divided  by  the  depth  of  the  truss.  Now  if  the 
pole  distance  in  thousands  of  pounds  is  taken  equal  to  the  depth 
of  the  truss  or  to  some  multiple  of  the  depth,  then  the  intercept 
will  represent  to  scale  the  stress  in  the  member ;  provided  a  scale, 
which  may  be  determined  in  the  following  manner,  is  used  in 
measuring  the  intercept.  In  this  problem,  D  =  20 ;  and  suppose 
that  H  =  40  ooo,  and  that  the  linear  scale  of  the  truss  is  I  inch  = 
40  feet.  Then  the  intercepts  should  be  measured  to  a  scale  of 

40  X —    — =80000  pounds  to  the  inch.    If  H  had  been  taken 

equal  to  20  ooo  pounds,  then  the  scale  to  be  used  in  measuring  the 
intercepts  should  have  been  I  inch  =  40  ooo  pounds. 

The  stresses  in  the  lower  chord  members  are  mean  propor- 
tionals between  those  in  the  adjacent  upper  chord  members,  and 
the  intercepts  are  to  be  taken  as  shown  in  Fig.  124,  b.  It  is  thus 
seen  that  the  stress  in  any  chord  member  may  be  found  directly 
by  scaling  the  intercept  under  the  center  of  moments. 

The  ordinates  to  the  stress  polygon  shown  in  Fig.  124,  b  may 
be  obtained  without  drawing  the  force  and  funicular  polygons, 
in  the  following  manner.  The  stress  in  the  center  member  X-6 
of  the  upper  chord  is  equal  to  the  bending  moment  at  the  center 
of  the  truss  due  to  the  uniform  load  of  w  Ibs.  per  linear  ft.  of 

truss  divided  by  the  depth  of  the  truss,  i.  e.,  to Lay  off  the 

8D 


240  STRESSES  IN  BRIDGE  TRUSSES.        Chap.  XXI. 

8oO  "X'  1 2O  'X'  1 2 

middle  ordinate  X-6  equal  to =  72  ooo,  to  any 

8  X  20 

given  scale.  Draw  the  horizontal  line  4-4  equal  to  one-half  the 
span  of  the  truss;  also  draw  the  line  1-4  parallel  to  the  middle 
ordinate  X-6.  Divide  the  horizontal  line  4-4  into  three  equal 
parts  (one-half  the  number  of  panels  in  the  truss)  ;  also  divide 
the  line  1-4  into  the  same  number  of  equal  parts,  and  number  as 
shown.  Draw  the  radial  lines  1-4,  2-4,  and  3-4  to  meet  verticals 
through  the  lower  chord  panel  points.  Then  these  points  of  inter- 
section will  give  points  on  the  required  stress  polygon. 

If  the  truss  has  an  odd  number  of  panels,  the  above  method 
should  be  modified  as  follows:  Divide  the  horizontal  line  4-4 
(one-half  the  span  of  the  truss)  into  as  many  parts  as  there  are 
panels  in  the  entire  truss ;  and  use  only  the  alternate  points  of 
division. 

It  is  seen  that  the  method  of  graphic  moments  and  shears  does 
not  give  the  kind  of  stress.  If  the  kind  of  stress  is  not  evident 
by  inspection,  then  it  may  be  found  by  algebraic  methods. 

185.  Live  Load  Chord  Stresses.    The  live  load  is  equal  to 
1600  Ibs.  per  ft.  per  truss,  and  the  live  panel  load  is  1600  X  20  = 
32000.     The  live  load  chord  stresses  may  be  obtained  by  con- 
structing a  diagram  similar  to  that  used  for  the  dead  load  chord 
stresses,  or  they  may  be  more  easily  obtained  from  the  dead  load 
stresses  by  direct  proportion. 

186.  Dead   Load   Web   Stresses.     The   reactions   for   the 
truss  loaded  with  the  dead  load  are  represented  by  Rx  and  R2 
(Fig.  124,  c).    In  a  bridge  with  parallel  chords,  the  web  members 
resist  the  shear.     The  shear  in  panel  L0Lj  is  equal  to  +  RI  >  m 
L,L2,  to  +  (R!  — W)  ;  and  in  L2L3,  to  +  (Rx  —  2\¥).  At  L3,  the 
shear  passes  through  o.     The  dead  load  shear  line  is  shown  by 
the  stepped  line  in  Fig.  124,  c.    Now  the  stresses  in  the  web  mem- 
bers are  equal  to  the  shears  in  the  members  multiplied  by  the 
secant  of  the  angle  that  the  members  make  with  the  vertical.    The 
shears  are  graphically  multiplied  by  the  secant  of  the  angle  by 
drawing  the  lines  cct,  ee^  and  ggt  parallel  to  1-2,  3-4,  and  5-6, 
respectively.    The  stresses  in  1-2,  3-4,  and  5-6  are  tension.    The 


Art.  3. 


GRAPHIC    MOMENTS    AND    SHEARS. 


241 


stresses  in  X-i,  2-3,  and  4-5  are  numerically  equal  to  those  in 
1-2,  3-4,  and  5-6,  respectively,  and  are  compression.  Since  the 
truss  and  loads  are  symmetrical  about  the  center,  it  is  only  neces- 
sary to  draw  the  diagram  for  one-half  of  the  truss. 

187.  Live  Load  Web  Stresses.  Another  diagram  must  be 
drawn  for  the  live  load  web  stresses.  This  diagram  may  be  con- 
structed in  the  following  manner : 

Assume  that  the  Warren  truss  shown  in  Fig.  125,  a  instead 
of  being  a  simple  span  is  a  cantiliver  truss,  the  right  end  being 

X 


|R,          f       _Jr_     Y^_ 
A~~    ~~ftr~    ~~frr~    ~~ft~~    ~~ftr~    ~~fi 

/   \  2   /    \  4   /    \  6  /    \  4'  /    \  r  / 
1  i    \       /  3    \     /  5  \     /  5'  \       /  3'  \      /I- 


\/ 


\    ' 

''' 


H=L 


D*ZOft-,  1  =  20 ft-;  L=l20ft;  W-16000  lbs-;P=32000  \bs- 
FIG.  125.     LIVE  LOAD  WEB  STRESSES — MAXIMUM  WEB  STRESSES. 

fixed  and  the  left  end  free.  Then  with  the  cantilever  truss  fully 
loaded  with  live  load,  lay  off  the  load  line  (Fig.  125,  c),  assume  a 
pole  O,  and  with  a  pole  distance  H  =  span  L,  draw  the  funicular 
polygon,  starting  the  polygon  at  a,  the  top  of  the  load  line.  Now 


242  STRESSES    IX    BRIDGE    TRUSSES.  Chap.  XXL 

the  bending  moment  at  the  right  support  is  equal  to  the  intercept 
Y!  multiplied  by  the  pole  distance  H.  But  the  given  truss  instead 
of  being  a  cantilever  truss  is  a  simple  span ;  and  is  held  in  posi- 
tion by  the  reaction  R±  at  the  left  end,  instead  of  being  fixed  at 
the  right  end.  Therefore  the  moment  at  the  right  end,  which  is 
equal  to  Hy1?  must  be  equal  to  the  moment  of  the  left  reaction. 
4  Now  the  moment  of  the  left  reaction  is  equal  to  RXL,  where  L  is 
the  span  of  the  truss ;  therefore  Hyt  =  RXL,  and  since  H  —  -  L  by 
construction,  y^  =  Rx. 

The  maximum  live  load  shear  in  X-i  and  1-2  is  obtained 
when  the  truss  is  fully  loaded  with  live  load,  and  is  yx  =  Rx.  The 
stress  in  X-i,  which  is  numerically  equal  to  that  in  1-2,  is 
obtained  by  drawing  a  line  through  a  point  a  parallel  to  the 
member  X-i. 

Now  move  the  truss  one  panel  to  the  left,  but  let  the  loads 
remain  stationary.  The  new  position  of  the  truss,  together  with 
the  new  loading,  is  shown  in  Fig.  125,  b.  With  the  same  pole  O 
and  pole  distance  H,  draw  a  new  funicular  polygon.  This  funicular 
polygon  will  coincide  with  a  part  of  the  first  polygon,  the  part  to 
the  left  of  the  point  d  (Fig.  125,  c),  being  identical  in  both  poly- 
gons. As  above,  the  bending  moment  at  the  right  end  for  this 
loading  is  equal  to  the  moment  of  the  left  reaction,  i.  e.,  Hy2  == 
R3L,  or  y2  =  R3.  Now  the  loading  shown  in  Fig.  125,  c  is  that 
for  a  maximum  shear  in  the  members  2-3  and  3-4;  and  since 
y2  =  R3  is  the  shear  in  these  members,  the  stresses  are  obtained 
by  drawing  through  d  a  line  parallel  to  the  member  2-3. 

In  like  manner,  gh  may  be  shown  to  be  the  shear  in  the  mem- 
bers 4-5  and  5-6  when  there  is  a  maximum  stress  in  4-5  and  5-6 ; 
and  the  stresses  in  these  members  are  obtained  by  drawing  a  line 
through  g  parallel  to  the  member  4-5. 

Likewise,  jm  is  equal  to  the  shear  in  6-5'  and  5 '-4'  when 
there  is  a  minimum  shear  in  6-5'  and  5 '-4',  and  the  stresses  in 
these  members  are  obtained  by  drawing  a  line  through  j  parallel 
to  the  member  6-5'. 

Similarly,  the  minimum  stresses  in  4'-3'  and  3'-2'  are  obtained 
by  drawing  a  line  through  o  parallel  to  4'~3'. 


Art.  4.  BOWSTRING    TRUSS.  243 

The  minimum  live  load  stresses  in  2'-!'  and  X'-i'  are  zero, 
when  there  is  no  live  load  on  the  truss. 

The  kind  of  stress  is  not  given  by  this  graphic  construction, 
and  if  not  evident  by  inspection,  it  may  be  obtained  by  algebraic 
methods. 

188.  Maximum  and  Minimum  Dead  and  Live  Load 
Stresses,  (a)  Chord  Stresses.  The  maximum  chord  stresses 
may  be  obtained  by  adding  the  dead  and  live  load  stresses.  The 
minimum  chord  stresses  are  equal  to  the  dead  load  stresses. 

(b)  Web  Stresses.  The  maximum  and  minimum  web  stresses 
may  be  obtained  by  graphically  combining  the  dead  and  live  load 
shears,  provided  the  dead  and  live  load  shear  diagrams  are  placed 
as  shown  in  Fig.  125,  c.  In  this  figure,  the  dead  load  shear  in 

X-i  is  be  =-  W,  the  live  load  shear  is  R1=y1  — ab,  and  the 

2 
maximum  shear  is  ac.  Likewise,  df  =  maximum  shear  in  2-3,  gi  = 

maximum  shear  in  4-5,  jk  =  minimum  shear  in  6-5',  no  =  mini- 
mum shear  in  4/~3/,  and  rs  =  minimum  shear  in  2'-!'.  The 
stresses  in  the  members  are  obtained  by  graphically  resolving  these 
shears. 


ART.  4.     STRESSES  IN  A  BOWSTRING  TRUSS — TRIANGULAR  WEB 

BRACING. 

The  solution  of  the  maximum  and  minimum  stresses  in  a  bow- 
string truss  will  be  taken  up  in  this  article.  The  dead  load  stresses 
will  be  determined  by  graphic  resolution,  and  the  live  load  stresses 
by  a  special  application  of  the  method  of  graphic  resolution.  The 
method  used  is  general  in  its  application  to  trusses  with  triangular 
web  bracing,  and  the  upper  chord  panel  points  may  lay  on  the  arc 
of  any  curve. 

189.  Problem.  It  is  required  to  find  the  maximum  and 
minimum  stresses  in  the  bowstring  truss  shown  in  Fig.  126,  a. 
The  upper  chord  panel  points  lay  on  the  arc  of  a  parabola  with  a 
middle  ordinate  of  24  ft.  The  truss  has  a  span  of  160  ft.,  and  a 


244 


STRESSES    IN    BRIDGE    TRUSSES. 


Chap.  XXI. 


panel  length  of  20  ft.    The  dead  load  will  be  taken  at  500  Ibs.  per 
ft.  per  truss,  and  the  live  load  at  1000  Ibs.  per  ft.  per  truss. 


R|+6ZJ   j+64.7    j+65.3   j  +65.5   j    (a) 


(b) 

Dead  Load 
Stress  Diagram 


FIG.  126.     DEAD  LOAD  STRESSES — BOWSTRING  TRUSS. 

190.  Chord  Stresses,  (a)    Dead  Load  Chord  Stresses.     The 
dead  panel  load  is  equal  to  500  X  20  =  10  ooo  Ibs.   The  dead  load 
chord  stresses  are  obtained  by  drawing  the  dead  load  stress  dia- 
gram for  one-half  of  the  truss,  as  shown  in  Fig.   126,  b.     The 
stresses,  expressed  in  thousands  of  pounds,  are  shown  on  the 
members  .of  the  truss  in  Fig.  126,  a. 

(b)  Live  Load  Chord  Stresses.  The  live  panel  load  is  equal 
to  1000  X  20  =  20  ooo  Ibs.  The  live  load  chord  stresses  are 
obtained  by  proportion  from  the  dead  load  chord  stresses.  These 
stresses  are  double  the  corresponding  dead  load  stresses. 

191.  Web  Stresses,     (a)    Dead  Load  Web  Stresses.     The 
dead  load  web  stresses  are  obtained  from  the  dead  load  stress 
diagram  shown  in  Fig.  126,  b.    These  stresses  are  shown  on  the 
members  of  the  truss  in  Fig.   126,  a.     It  is  seen  that  the  web 
stresses  in  this  type  of  truss  are  small,  and  that  the  stresses  in 


Art.  4. 


BOWSTRING   TRUSS. 


245 


all  the  web  members  are  tension.    These  members  act  as  auxiliary 
members  to  transfer  the  load  to  the  upper  chord  panel  points. 

(b)  Live  Load  Web  Stresses.  The  live  load  web  stresses 
are  obtained  from  the  stress  diagram  shown  in  Fig.  127,  b.  This 
diagram  is  constructed  by  assuming  a  value  (in  this  case  10000 
Ibs.)  for  the  left  reaction  Rx,  and  drawing  the  stress  diagram 
with  no  loads  on  the  truss. 

X 


R, 


Panel  Load  =20000  Ibs. 


Member 

True  Ri 

Assumed  Ri 

1-2 
2-3,  3-4 
4-5,  5-6 
6-7,  7-8 
7J-8,  6'-7' 
5-6',  4-5' 
3'-4',  2-3' 

r-21 

7.00 
5-25 
3-75 
2.50 
1-50 
0.75 
0.25 
0 

(a)     Uy          ^3  Ll*  L'I  ~|^« 

I    Stress  Diagram  for  |  R* 

Max- and  Min-Live  Load  Stresses  in  Web  Members. 
Ri  assumed -No  loads  on  truss. 

"To  cjet  stress  in  any  web  member, 

mul.  measured  stress  by        ue  ^' 


Assumed  Ri 
20000**    40000* 


(O 
FIG.  127. 


LIVE  LOAD  WEB  STRESSES — BOWSTRING  TRUSS. 


It  has  already  been  shown  that,  for  a  maximum  live  load  stress 
in  any  web  member,  the  longer  segment  of  the  truss  should  be 
loaded ;  and  for  a  minimum  live  load  stress,  the  shorter  segment 
should  be  loaded.  Therefore,  for  a  maximum  stress  in  any  web 
member  to  the  left  of  the  center  of  the  truss,  there  should  be  no 
loads  to  the  left  of  that  member;  and  for  a  minimum  stress  in 
the  corresponding  member  on  the  right  of  the  center,  there  should 
be  no  loads  to  the  left  of  that  member.  The  shear  in  any  member 
is  therefore  equal  to  the  reaction ;  and  the  stress  in  the  member 
may  be  found  by  multiplying  the  stress  in  that  member,  obtained 


246  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

from  the  stress  diagram  shown  in  Fig.  127,  b,  by  the  ratio  of  the 
actual  live  load  reaction  (for  a  maximum  or  minimum  stress  in 
that  member)  to  the  assumed  reaction.  For  example,  to  get  a 
maximum  stress  in  1-2,  the  bridge  is  fully  loaded  with  live  load, 
and  the  reaction  1^=70  ooo.  The  maximum  stress  in  1-2  is  there- 
fore equal  to  the  stress  scaled  from  Fig.  127,  b  multiplied  by 

7^ 

IOOOO 

For  stresses  in  2-3  and  3-4,  load  all  joints  except  'Li ;  for 
stresses  in  4-5  and  5-6,  load  all  joints  except  Lt  and  L2;  for 
stresses  in  6-7  and  7-8,  load  all  joints  except  Lx,  L2,  and  L3 ;  for 
stresses  in  8-7'  and  7'-6',  load  the  joints  L3',  L2',  and  L/;  for 
stresses  in  6/~5/  and  5'-4',  load  joints  L2'  and  L/ ;  for  stresses  in 
4'~3'  and  3'-2',  load  the  joint  L/  only;  and  for  stress  in  2'-!', 
there  should  be  no  loads  on  the  truss. 

The  ratios  of  the  actual  reactions  to  the  assumed  reactions  for 
the  above  loadings  are  shown  in  Fig.  127,  c. 

The  live  load  stresses  are  shown  on  the  members  of  the  truss 
in  Fig.  127,  a,  and  are  obtained  by  scaling  the  stresses  from  the 
diagram  in  Fig.  127,  b  and  multiplying  these  stresses  by  the  ratios 
shown  in  Fig.  127,  c. 

192.  Maximum  and  Minimum  Dead  and  Live  Load 
Stresses,  (a)  Chord  Stresses.  The  maximum  chord  stresses  are 


+  186.3  -H94.I  +195.9  +196-5 

•*•  6Z.I  +  64.7  +  65.3  +  65-5 

FIG.  128.     MAXIMUM  AND  MINIMUM  STRESSES — BOWSTRING  TRUSS. 

equal  to  the  sums  of  the  corresponding  dead  and  live  load  stresses, 
and  the  minimum  chord  stresses  are  the  dead  load  stresses.  These 
stresses  are  shown  on  the  truss  diagram  in  Fig.  128. 


Art.  5.  PARABOLIC    BOWSTRING    TRUSS.  247 

(b)  IV eb  Stresses.  The  maximum  and  minimum  web  stresses 
are  obtained  by  combining  the  dead  and  live  load  stresses  shown 
in  Fig.  126,  a  and  Fig.  127,  a,  respectively.  In  making  the  com- 
binations, both  the  stresses  in  the  member  and  those  in  the  corre- 
sponding member  on  the  other  side  of  the  center  line  of  the  truss 
must  be  considered. 

The  maximum  stress  in  1-2  is  +7.9+15.8  =  4-23.7,  and 
the  minimum  stress  is  +  7.9,  the  dead  load  stress. 

The  maximum  stress  in  2-3  is  +  7.3  +  21.6  —  +  28.9,  and  the 
minimum  stress  is  +  7.3  —  6.8  =  +  0.5. 

The  maximum  stress  in  3-4  is  +  5.5  +  20.5  =  +  26.0,  and  the 
minimum  stress  is  +  5.5  —  9.1  =  —  3.6. 

The  maximum  stress  in  4-5  is  +  6.0  +  25.3  =  +  31.3,  and  the 
minimum  stress  is  +  6.0 —  13.2  =  — 7.2. 

The  maximum  stress  in  5-6  is  +  5.3  +  24.7  =  +  30.0,  and 
the  minimum  stress  is  +  5.3  —  13. 7  =  —  8.4. 

The  maximum  stress  in  6-7  is  +5.2  +  27.0  =  +  32.2,  and 
the  minimum  stress  is  +  5.2  —  16.2  =  —  i i.o. 

The  maximum  stress  in  7-8  is  +  5.4  +  27.0  =  +  32.4,  and  the 
minimum  stress  is  +  5.4 —  16.2  =  —  10.8. 


ART.  5.     STRESSES  IN  A  PARABOLIC  BOWSTRING  TRUSS. 

The  maximum  and  minimum  stresses  in  a  parabolic  bowstring 
truss  will  be  found  in  this  article  by  a  special  graphic  method. 
The  method  given  applies  only  to  trusses  whose  upper  chords 
panel  points  lay  on  the  arc  of  a  parabola. 

193.  Problem.     It  is  required  to  find  the  maximum  and 
minimum  stresses  in  the  parabolic  bowstring  truss,  half  of  which 
is  shown  in  Fig.  129.     The  truss  has  a  span  of  160  ft.,  a  panel 
length  of  20  ft.,  and  a  depth  at  the  center  of  24  ft.    The  dead  load 
will  be  taken  at  500  Ibs.  per  ft.  per  truss,  and  the  live  load  at  1000 
Ibs.  per  ft.  per  truss. 

194.  Chord  Stresses,  (a)    Dead  Load  Chord  Stresses.    The 
dead  panel  load  =  500  X  20=  10000  Ibs.,  or  say  10.0.    Now  the 
bending  moment  at  any  point  in  a  truss  loaded  with  a  uniform  load 


248  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

varies  as  the  ordinates  to  a  parabola ;  and  the  stress  in  any  chord 
member  is  equal  to  the  bending  moment  divided  by  the  depth  of 
the  truss.  Since  the  moment  at  any  point  varies  as  the  ordinates 
to  a  parabola,  and  the  moment  arm  for  the  stress  in  any  lower 


a1 1 

FIG.    129.     MAXIMUM   AND  MINIMUM    STRESSES — PARABOLIC   BOWSTRING  TRUSS. 

chord  member  is  an  ordinate  to  the  parabola,  it  is  seen  that  the 
stress  in  the  lower  chord  is  constant. 

For  this  truss,  the  moment  at  the  center  = ,  and  the  dead 

wL2          500  X  1 60  X  1 60 
load   stress   in  the   lower   chord  =    -j=-  ==    ^ 

=  66  700.  Since  the  stress  in  the  lower  chord  is  constant  when  the 
bridge  is  fully  loaded,  it  is  seen  that  the  diagonals  are  not  stressed, 
and  that  the  horizontal  components  of  the  stresses  in  the  upper 
and  lower  chords  are  equal.  The  dead  load  stresses  in  the  upper 
chord  members  may  be  found  graphically,  as  follows :  Lay  off  ab 
(Fig.  129)  equal  to  the  stress  in  the  lower  chord,  and  draw  the 
vertical  lines  aa'  and  bb'.  Draw  lines  parallel  to  the  upper  chord 
members,  and  the  distances  intercepted  on  these  lines  between  the 
vertical  lines  aa'  and  bb'  will  represent  the  stresses  in  the  corre- 
sponding upper  chord  members. 


Art.  5.  PARABOLIC    BOWSTRING   TRUSS.  249 

(b)  Live  Load  Chord  Stresses.  The  maximum  live  load  chord 
stresses  are  obtained  when  the  bridge  is  fully  loaded  with  live  load, 
and  are  found  in  the  same  manner  as  the  dead  load  ch'ord  stresses. 
To  find  these  stresses,  lay  off  be  (Fig.  129)  equal  to  the  live  load 

1000  X  1 60  X  1 60 

stress  in  the  rower  chord  = ^— =  133  300.      The 

o  X  24 

upper  chord  stresses  are  represented  by  the  distances  intercepted 
between  the  vertical  lines  bb'  and  cc'  on  lines  drawn  parallel  to 
the  chord  members. 

The  minimum  live  load  chord  stresses  are  zero. 

195.  Web  Stresses,  (a)  Dead  Load  Web  Stresses.  Since 
the  stress  in  the  lower  chord  is  constant  when  the  bridge  is  loaded 
with  a  uniform  load,  it  is  seen  that  the  dead  load  stresses  in  the 
diagonals  are  zero.  The  dead  load  stresses  in  the  verticals  are 
tensile  stresses,  and  are  each  equal  to  the  dead  panel  load 
=  +  10.0. 

(b)  Live  Load  Web  Stresses.  The  maximum  live  load 
stresses  in  the  diagonals  and  the  minimum  live  load  stresses  in 
the  verticals  may  be  obtained  directly  from  the  diagram  shown  in 
Fig.  129.  The  maximum  live  load  stresses  in  the  verticals,  and 
the  minimum  live  load  stresses  in  the  diagonals  are  obtained  when 
the  bridge  is  fully  loaded  with  live  load.  The  maximum  live  load 
stresses  in  the  verticals  are  +  2O-°  (tne  n've  panel  load),  and  the 
minimum  live  load  stresses  in  the  diagonals  are  o. 

The  diagram  shown  in  Fig.   129  is  constructed,  as  follows: 

1000  X  1 60  X  1 60 

Lay  off  be  to  any  scale=  -        g =  133  300.  Construct 

o  X  24 

the  half  truss  diagram,  making  the  half  span  equal  to  b'd'  =  Jbc. 
Since  the  span  is  160  ft.  and  the  depth  of  the  truss  is  24  ft.,  make 

dd'  =— —  be.    With  b'd'  equal  to  the  half  span  and  dd'  equal  to 

1 OO 

the  depth  at  the  center,  draw  the  outline  of  the  given  truss. 

The  maximum  live  load  stresses  in  the  diagonals  sloping  upward 
to  the  left  are  obtained  when  the  longer  segment  of  the  bridge  is 
loaded,  and  the  maximum  stresses  in  the  diagonals  sloping  upward 


250  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXL 

to  the  right,  when  the  shorter  segment  of  the  bridge  is  loaded  with 
live  load.  These  stresses  are  obtained  directly  by  scaling  the  mem- 
bers to  the  given  scale.  The  minimum  live  load  stress  in  the  first 
vertical  from  the  left  end  is  zero.  The  minimum  live  load  stresses 
in  the  other  verticals  are  compression,  and  are  represented  by  the 
vertical  distances  between  the  first  upper  chord  panel  point  and 
each  succeeding  upper  chord  panel  point. 

The  proof  of  the  construction  shown  in  Fig.  129  is  given 
in  §  197. 

196.  Maximum    and    Minimum    Dead    and    Live    Load 
Stresses,     (a)     Chord  Stresses.     The   maximum   stress   in   the 
lower  chord  is  represented  by  ac   (Fig.   129),  which  represents 
the  sum  of  the  dead  and  live  load  chord  stresses.    The  maximum 
stress  in  the  upper  chord  members  are  represented  by  the  dis- 
tances intercepted  between  the  vertical  lines  aa'  and  cc'  by  the 
upper  chord  members  produced.    The  minimum  chord  stresses  are 
the   dead   load    stresses.      The   maximum   and   minimum   chord 
stresses  are  shown  on  the  members  produced. 

(b)  Web  Stresses.  The  maximum  stresses  in  the  diagonals 
are  represented  by  the  lengths  of  the  diagonals,  the  dead  load 
stresses  in  these  members  being  zero.  These  stresses  are  shown 
on  the  members  of  the  truss  diagram.  The  minimum  stresses  in 
the  diagonals  are  zero..  The  maximum  stresses  in  the  verticals 
are  equal  to  the  sums  of  the  dead  and  live  panel  loads  =  +  10.0  + 
20.0  =  +  30.0.  The  minimum  stresses  in  the  verticals  are  shown  in 
Fig.  129.  These  stresses  are  obtained  by  laying  off  to  scale  the  dead 
panel  load,  W=io.o,  above  the  first  upper  chord  panel  point, 
and  drawing  the  horizontal  dot  and  dash  line  shown  in  Fig.  129. 
The  dead  and  live  load  stresses  are  of  an  opposite  kind,  and  the 
resulting  stress  is  thus  obtained.  The  minimum  stresses  in  the 
verticals  are  represented  by  the  distances  between  this  line  and 
each  upper  chord  panel  point.  Distances  measured  below  the  line 
represent  tensile  stresses,  and  those  above,  represent  compressive 
stresses. 

197.  Proof  of  Construction  Shown  in  Fig.   129.     It  will 
now  be  proved  that   for  the  construction  shown  in  Fig.  129,  (a) 
the  maximum  live  load  stresses  in  the  diagonals  are  represented  by 


Art.  5. 


PARABOLIC    BOWSTRING    TRUSS. 


251 


the  lengths  of  the  diagonals,  and  (b)  that  the  minimum  live  load 
stresses  in  the  verticals,  i.  e.,  the  maximum  live  load  compressive 
stresses,  are  represented  by  the  vertical  distances  from  the  first 
upper  chord  panel  point  to  each  upper  chord  panel  point. 

U3 


Ls 


FIG.  130.     PARABOLIC  BOWSTRING  TRUSS. 


(a)    Stresses  in  Diagonals.     Let  Fig.  130  represent  a  parabolic  bow- 
string   truss    in    which    the    span  —  L ,    and    the    depth =  _  X  J- = 

L-L  —  panel  load  P. 

8 

Also,  let  x  =  abscissa  and  y  =  ordinate  of  any  point  O  on  the  parabola, 
p=: live  .load  per  foot, 
n  =  number  of  panels  in  truss, 
m  =  number  of  loads  on  truss, 
L  —  span  of  truss, 
D  =  depth  of  truss. 

It  is  required  to  prove  that  the  maximum  live  load  stress  in  any 
diagonal  is  represented  by  the  length  of  that  diagonal. 
The  equation  of  the  parabola  about  the  left  .end  L0  is 


from  which, 


x). 


Now  the  horizontal  component  of  the  maximum  live  load  stress  in 
any  diagonal  member  is  equal  to  the  difference  of  stress  in  the  two  adja- 
cent lower  chord  members  when  the  truss  is  loaded  for  a  maximum  stress 
in  that  diagonal. 

Consider  any  diagonal,  say  U2L3  in  panel  L2L3,  whose  maximum  stress 
occurs  when  there  are  m  loads  on  the  truss. 


_,       ,,  .     .      ...         .p. 

For  this  loading,  E, 

fe' 


(m  +  l)m 
-  ' 


Jfii 

n 


Moment  of  T?t  about  the  panel  point  to  right  of  diagonal  = 

(m  -f-  l)m       rL        /„       —.x  L 
__ , —    . .  y  ___  y  (n  —  m)  — , 

2n         X    u   X  ^  n 


252  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

and  stress  in  lower  chord  member  to  right  of  diagonal  = 

(m  +  l)m        pL        (n_m)L 
2n          X    n  X  n 


pL4(m  +  1)    (n  —  m)rn 
8nyDx  (L  —  x) 

In  above  equation,  x  =  _  (n  —  m),  and  substituting  this  value  of  x, 

n 
we  have 

pL4  (m  -jr-  1)  (n  —  m)m 

S  — Y 


8n3D^(n  — m)  [L—  ~(n  —  m)] 
n  n 

pL2  (m  +  1) 


8nD 

Also  moment  of  E!  about  panel  point  to  left  of  diagonal  = 
(m  4-  l)m        pL         /  -.N   L 


and  stress  in  lower  chord  member  in  same  panel  as  diagonal  = 


(m  +  l)m        pL  .    L 

--x'-x  (»—  -i) 


pL4(m  +  l)    (n  —  m  — 


8naDx  (L  —  x) 

equation,    x  = 
of  x,  we  have 


In    above    equation,    x=r~:!(n  —  m — 1),    and   substituting    this    value 


pL4  (m  +  1)  (n  —  m  —  l)m 


8n3D--t(n  —  m  —  1)  [L  —  t  (n  —  m  —  1)] 
n  n 


pL2m 
~8nD~ 


Now  S  —  S'  = 


, pL2  (m+ !)•        pLrm 


8nD  "    8nD 


—  !? =  horizontal  component  of  stress  in  diagonal. 

8nD 

Now  the  span  of  the  truss  was  laid  off  equal  to  1^ —  .     Therefore 

81) 
the  horizontal  component  of  the  maximum  stress  in  the  diagonal,  which  is 

— — ,  is  represented  by  one  panel  length  of  the  lower  chord.     Since  one 
panel  length   represents   the   horizontal   component   of  the   stress   in  the 


Art.  5.  PARABOLIC    BOWSTRING    TRUSS.  253 

diagonal,  the  length  of  the  diagonal  itself  will  represent  the  maximum 
stress  in  it. 

(b)  Stresses  in  Verticals.  It  is  required  to  prove  that  the  minimum 
live  load  stresses  in  the  verticals,  i.  o.,  the  maximum  compressive  stresses, 
are  represented  by  the  vertical  distances  from  the  first  upper  chord 
panel  point  to  each  upper  chord  panel  point. 

Consider  any  vertical,  say  U2L,  (Fig.  130),  whose  maximum  com- 
pressive stress  occurs  when  there  are  m  loads  on  the  truss.  Now  the 
stress  in  U2L2  when  there  are  m  loads  on  the  truss  is  equal  to  the 
difference  in  stress  between  L2L:,  and  LtL2  for  this  loading  multiplied 
by  the  tangent  of  the  angle  that  t^L,,  makes  with  LjL2. 

The  stress  in  the  lower  chord  member  to  the  right  of  the  vertical  is 

pL2m 
'    8nD  ' 

The  stress  in  the  lower  chord  member  to  the  left  of  the  vertical  is 

K!  (n  —  m  —  2)  - 

n 


x.^x.(a-«-l).| 


pL4(m  4-  1)  (n  —  m  —  2)m 
8n3Dx  (L  —  x) 

In    above    equation  x=  —  -(n  —  m  —  2),  and  substituting  this  value 
of  x,  we  have 

,,_  pL4(m  +  l)(n  —  m  —  2)m 


8nT>-  <n  —  m  —  2)[L—  i  (n  —  m  —  2)] 
n  n 


_  pL2(m-f 

=   8nD(m  +  2j  * 

—  pL2m  _  pL2(m-f-l)m 
8nD    "      8nD(m  +  2) 

pL2m 


~  8nD(m  +  2) 
The  stress  in  any  vertical  member  is 

- — tan    0,    where    6    is   the    angle    that    the    diagonal 

SnD(m  +  2) 

which   meets   the   required   vertical   at   the   lower  chord  makes  with  the 
lower    chord. 


254  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

An  expression  will  now  be  found  for  tan  9. 

Tan  e  =  L  =  H7. 

L         L 

n 

Now  y=    4Dx    -(L  —  x),  in  which  x  =   ^  -(n  —  m  —  2). 
L" 

Substituting  values  of  x  and  y,  we  have 


x 

^ 


(n  —  m  —  2)[L—         <n  —  m  —  2)] 


1    •>         y\.  *  /   L 

L-     ^  n  n 

tan  O  =  — 


L 

4D(n  — m  — 2)(m  -f-  2) 


Stress  in  vertical  ~ 


Ln 

pL2m  4D(n  — m  —  2)(m  +  2) 


8nD(m-[-2)    *  Ln 


pLm(n-m-2)  pnlm(n-m-2)     ?  where  l  =  panel  lengtll? 


n~ 


Pm(n  —  m  —  2) 
—  —     -       —  ,  where  P  =  panel  load. 

For  the  problem  given  in  §  193,  where  n  =  8; 

when  m  —  6,  then  stress  in  vertical  =  o; 


Since,  by  construction,  D=  ^^  ?  _  =  "P,  it  is  seen  that  the  above 

L          8JJ 

stresses  in  the  verticals  are  represented  by  the  ordinates  to  a  parabola 
whose  ordinates  are  igP  less  than  those  to  the  first  parabola. 

The  above  relation  may  be  proved  in  a  different  manner,  as 
follows  :  Produce  each  upper  chord  member  of  the  eight-panel 
truss  (Fig.  130)  to  intersect  the  lower  chord  produced.  Load 
the  bridge  for  the  maximum  compressive  stress  in  each  vertical, 
and  find  the  stresses  in  the  verticals  by  moments.  If  1  represents 
a  panel  length,  the  moment  arms  for  the  members  UjLj,  U2L2, 
U3L3,  and  U4L4  are  1,  2  f  1,  5  1,  and  16  1,  respectively  ;  the  reactions 
for  maximum  compressive  stresses  are  ^  P,  *£-  P,  -^P,  and  f  P; 
and  the  stresses  are  o,  ^%-P,-^  P,  and^-P.  Now  the  above  ordi- 
nates are  equal  to  the  ordinates  to  the  given  parabola  minus  the 
first  ordinate 


Art.  6.  WIND  LOAD  STRESSES  IN   LATERALS.  255 

ART.  6.     WIND  LOAD  STRESSES  IN  LATERAL  SYSTEMS. 

The  wind  load  stresses  in  the  upper  and  lower  laterals  of  a 
bridge  will  be  determined  in  this  article.  All  the  wind  load  will.be 
assumed  to  act  on  the  windward  truss ;  although  the  assumption 
is  sometimes  made  that  half  of  the  wind  acts  on  each  truss.  An 
initial  tension  is  sometimes  put  into  the  laterals  to  give  additional 
rigidity  to  the  truss,  but  will  not  be  considered  in  this  article. 
The  upper  chord  members  are  a  part  of  the  top  lateral  system,  and 
the  lower  chord  members  and  floorbeams  are  a  part  of  the  bottom 
lateral  system.  Most  specifications  state  that  the  wind  load 
stresses  need  not  be  considered  unless  they  are  25,  or,  in  some 
cases,  30  per  cent  of  the  live  and  dead  load  stresses. 

198.  Upper  Laterals.  It  is  required  to  find  the  stresses  in 
the  upper  lateral  system  shown  in  Fig.  131,  a.  The  panel  length 
is  20  ft.,  and  the  width  is  16  ft.  The  wind  load  will  be  taken  at 
150  Ibs.  per  ft.,  and  will  be  treated  as  a  fixed  load.  The  panel  load 
W=  150  X  20  =  3000  Ibs.  The  end  strut  of  the  upper  lateral 
system  is  also  a  part  of  the  portal  system,  and  its  stress  will  not  be 
considered  at  this  time. 


?>                               0           5000        10000 
3,3' i £ TI  X ,Y  l   ....  t | 


(b) 


Y       Fixed  Load 

Panel  Length  =  20',  Width  =  1 6'. 
v  W=  3000  Ibs. 

FIG.  131.     WIND  LOAD  STRESSES  IN  UPPER  LATERALS. 


(a)  Chord  Stresses.  The  fixed  wind  load  stress  diagram  for 
one-half  of  the  truss  is  shown  in  Fig.  131,  b.  When  the  wind  acts 
in  the  direction  shown,  the  diagonals  represented  by  full  lines  are 
in  action ;  and  the  stress  diagram  was  drawn  using  these  diago- 
nals. When  the  direction  of  the  wind  is  reversed,  the  diagonals 
shown  by  dotted  lines  are  in  action.  The  stresses  in  the  chord 


256 


STRESSES    IX    BRIDGE    TRUSSES. 


Chap.  XXI. 


members  for  the  wind  acting  in  the  direction  shown  in  Fig.  131,  a 
are  given  in  the  first  line  of  Fig.  132,  a.  The  stresses  in  these 
members  when  the  direction  of  the  wind  is  reversed  are  given  in 
the  second  line  of  Fig.  132,  a.  It  is  seen  that  the  chord  members 
are  subjected  to  reversals  of  stress  due  to  the  wind  load.  Since 


STRESSES  IN  CHORD  MEMBERS 

Chord  Member 

Y-l 

Y-2 

Y-3 

x-r 

X-Z1 

X-5' 

Wind  as  shown 
Opposite  direction 

-  7.50 
0.00 

-11.25 
+   7.50 

-  11.25 
-»-  11.25 

0.00 
-  7.50 

+   7.50 
-  11.25 

+  11.25 
-  11.25 

(a) 


STRESSES  IN  WEB  MEMBERS 

Web  Member 

r-i 

I-Z' 

2'-Z 

2-3' 

3L3 

Wind  as  shown 

+  9.60 

-6.00 

4-4-80 

-3.00 

0.00 

When  wind  comes  from  opposite  direction  .other  set  of  diagonals  acts. 

ibl 

PIG.  132.     TABLE  OP  STRESSES  —  UPPER  LATERALS. 

the  chord  members  are  already  stressed,  it  is  seen  that  there  are 
actually  no  reversals  of  stress  unless  the  wind  load  stresses  are  of 
an  opposite  kind  and  are  larger  than  the  stresses  already  in  the 
chord  members. 

(b)  Web  Stresses.  The  wind  load  stresses  in  the  web  mem- 
bers may  be  obtained  from  the  stress  diagram  shown  in  Fig. 
131,  b.  These  stresses  are  shown  in  Fig.  132,  b.  When  the  wind 
acts  from  the  opposite  direction,  the  set  of  diagonals  shown  by  the 
dotted  lines  is  in  action ;  and  the  stresses  in  them  are  equal  to  the 
stresses  in  the  corresponding  members  when  the  wind  acts  in  the 
direction  shown. 

199.  Lower  Laterals.  It  is  required  to  find  the  stresses  in 
the  lower  lateral  system  shown  in  Fig.  133,  a.  The  panel  length 
is  20  ft.,  and  the  width  is  16  ft.  The  wind  load  will  be  taken  at 
600  Ibs.  per  ft.  Of  this  load,  150  Ibs.  per  ft.  will  be  treated  as  a 
fixed  load,  and  450  Ibs.  per  ft.  as  a  moving  load.  The  fixed  panel 
load  is  150  X  20  =  3000  Ibs.,  and  the  moving  panel  load  is  450  X 
20  =  9000  Ibs. 

(a)  Chord  Stresses,  (i)  Fired  Load.  When  the  wind  acts 
in  the  direction  shown  in  Fig.  133,  a,  the  diagonals  shown  by  full 


Art.  6. 


WIND   LOAD   STRESSES    IN    LATERALS. 


257 


lines  are  in  action.  The  fixed  wind  load  stress  diagram  is  shown 
in  Fig.  133,  b;  and  the  stresses  in  the  chord  members  are  shown 
in  Fig.  134,  a. 

X 


(b)        ,„ , 

B  B1  (c) 

Fixed  Load  Moving  Load 

Panel  Length  =  20'-,  Width  =  16';  W  =  3000  lbs.,-P  =  9000  Ibs- 
FIG.  133.     WIND  LOAD  STRESSES  IN  LOWER  LATERALS. 

(2)  Moving  Wind  Load.  The  maximum  stresses  in  the  chord 
members  due  to  the  moving  load  are  obtained  when  the  load 
covers  the  entire  span.  Since  the  moving  panel  load  is  three  times 
that  of  the  fixed  load,  these  stresses  are  obtained  by  multiplying 
the  corresponding  fixed  wind  load  stresses  by  three.  The  stresses 
due  to  the  moving  load  are  given  in  the  second  line  of  Fig.  134,  a. 

(b)  Web  Stresses,  (i)  Fixed  Load.  The  stresses  in  the 
lower  chord  web  members  for  the  fixed  load  of  150  Ibs.  per  ft.  are 
obtained  from  the  stress  diagram  shown  in  Fig.  133,  b.  These 
stresses  are  given  in  Fig.  134,  b. 

(2)  Moving  Load.  The  stresses  in  the  lower  chord  web  mem- 
bers for  the  moving  load  of  450  Ibs.  per  ft.  are  obtained  from  the 
stress  triangle  shown  in  Fig.  133,  c.  This  triangle  is  constructed 
as  follows :  Lay  off  A'C  equal  to  twice  the  width,  and  B'C  equal 
to  twice  the  panel  length ;  and  draw  the  line  A'B'.  Then  lay  off 
AC  equal  to  the  moving  panel  load  P,  and  draw  AB  parallel  to 
A'B'  to  meet  the  horizontal  line  BC.  Then  AC  and  AB  repre- 
sent the  stresses  in  the  struts  and  diagonals,  respectively,  due  to 
a  reaction  equal  to  P.  The  stresses  in  these  members  due  to  any 


258 


STRESSES    IN    BRIDGE    TRUSSES. 


Chap.  XXI. 


reaction  are  obtained  by  proportion.  The  stresses  in  the  web 
members  to  the  left  of  the  center  of  the  truss  for  a  moving  load 
at  each  of  the  lower  chord  panel  points  are  shown  in  Fig.  134,  c. 
The  maximum  stress  in  each  member  due  to  the  moving  load  is 
shown  in  the  last  line  of  Fig.  134,  c. 


STRESSES    IN    CHORD     MEMBERS 

Chord  Member 

Y-l 

Y-2 

Y-3 

Y-4 

x-r 

X-21 

X-3' 

X-4' 

Fixed  Wind  Load 
Movinq  Wind  Load 
Maximum  Stress 
Opposite  direction 

-11.25 
-33.75 

-45.00 
-  0.00 

-18.75 
-56.25 
-75.00 
+45.00 

-22.50 
-67.50 
-90.00 
+75-00 

-22.50 
-67.50 
-90.00 
H-90.00 

0.00 
0.00 
0.00 
-45.00 

+  11.25 
+33.75 
+45.00 
-75.00 

+  18.75 
+56.25 
+75.00 
-90.00 

+22.50 
+67.50 
+90.00 
-90.00 

(a) 


STRESSES  IN  WEB  MEMBERS  -FIXED  LOAD 

Web  Member 

I'-l 

I-Z' 

2'-2 

2-3' 

3'-3 

3-4 

4'-4 

Fixed  Wind  Load 

+  14.40 

-9.00 

+  9.60 

-6.00 

+  4.80 

-3-00 

0-00 

(b) 


STRESSES  IN  WEB  MEMBERS  -MOVING  LOAD 

Web  Member 

I'-l 

1-2' 

2LZ 

2-3' 

3'-3 

3-4' 

4-4 

Moving  Load  at  \L\ 

+  2.06 

-  1.29 

+  2.06 

-   1.29 

+  2.06 

-  1.29 

+  2.06 

„          „     ^ 

+  4.12 

-  2.58 

+  4.12 

-  2.58 

+  4.12 

-2.58 

+  4.12 

»         ».     »  L^ 

+  6.18 

-  3.67 

+  6.18 

-  3.87 

+  6.18 

-3.67 

+  6.16 

n         n     [_3, 

+  8.24 

-  5.16 

+  8.24 

-  5.16 

+  8.24 

-5.16 

Lz 

+  10.30 

-6.45 

+  10.30 

-6.45 

„     „  Ll 

+  12.36 

-7.74 

Max-  Movinq  Load 

+43.26 

-27.09 

+30.90 

-19.35 

+20.60 

-12.90 

+  12.36 

(0 


MAXIMUM   STRESSES  IN  WEB  MEMBERS 

Web  Member 

r-i 

1-2' 

2'-2 

2-3' 

3'-3 

3-4' 

4'  -4 

Fixed  Wind  Load 
Movinq  Wind  Load 
Maximum  Stress 

+  14.40 
+43.26 
+  57.66 

-9.00 
-27.09 
-36.09 

+  9.60 
+30.90 
+40.50 

-  6.00 
-19.35 
-25.35 

+  4.80 
+  20.60 
+  25.40 

-  3.00 
-12.90 
-15.90 

0.00 
+12.36 
+  12.36 

When  wind  comes  from  opposite  direction  ,  other  set  of  diagonals  acts 

Id) 
FIG.  134.     TABLE  OF  STRESSES  —  LOWER  LATERALS. 

200.  Maximum  and  Minimum  Stresses  in  Upper  Laterals. 
The  upper  laterals  are  loaded  with  a  fixed  load  only,  and  the  maxi- 
mum chord  stresses  are  shown  in  the  first  line  of  Fig.  132,  a.  The 


Art.  7. 


METHOD    OF    COEFFICIENTS. 


259 


minimum  chord  stresses  are  obtained  when  the  wind  acts  from 
the  opposite  direction,  and  are  given  in  the  second  line  of  Fig. 
132,  a. 

The  maximum  web  stresses  are  shown  in  Fig.  132,  b.  The 
minimum  web  stresses  are  zero. 

201.  Maximum  and  Minimum  Stresses  in  Lower  Laterals. 
The  maximum  stresses  in  the  lower  chord  members  for  fixed  and 
moving  loads,  when  the  wind  acts  in  the  direction  shown,  are 
given  in  the  third  line  of  Fig.  134,  a.  The  minimum  chord  stresses 
are  obtained  when  the  wind  acts  from  the  opposite  direction,  and 
are  shown  in  the  fourth  line  of  Fig.  134,  a. 

The  maximum  web  stresses  for  fixed  and  moving  loads  are 
given  in  Fig.  134,  d.  When  the  wind  acts  from  the  other  direc- 
tion, the  other  set  of  diagonals  is  thrown  into  action,  and  the 
stresses  in  the  first  set  are  zero. 


ART.  7.     STRESSES  IN  TRUSSES  WITH  PARALLEL  CHORDS  BY  THE 
METHOD  OF  COEFFICIENTS. 


In  trusses  with  parallel  chords,  the  bending  moment  is  resisted 
by  the  chords,  and  the  shear  by  the  web  members.  For  such 
trusses,  the  method  explained  in  this  article  may  be  used  to  advan- 
tage for  determining  the  stresses. 

202.     Algebraic   Resolution — Method  of  Coefficients.     To 

U, 


6  '6  '6  '6  '  6          p 

Chord   Stresses  =  Coefficients  X  P  tan  0 
Web     Stresses  =  Coefficients  X  P  sec  9 

Coefficients   for  a  sinqle   load 
FIG.  135.     DEAD  LOAD  COEFFICIENTS — WARREN  TRUSS. 

explain  this  method,  let  it  be  required  to  find  the  stresses  in  the 
members  of  the  truss  shown  in  Fig.   135,  the  truss  being  first 


260  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

loaded  with  a  single  load  P.  For  this  loading,  the  left  reaction  is 
equal  to  -J-  P,  and  the  right  reaction  to  |-  P.  For  equilibrium, 
the  summation  of  both  the  horizontal  and  vertical  forces  at  any 
point  must  be  equal  to  zero.  Resolving  the  forces  at  L0,  it  is  seen 
that  the  stress  in  the  member  X-i  =  —  -J  P  sec  0,  and  the  stress 
in  Y-i  =  -f-  -J-  P  tan  0;  where  6  is  the  angle  that  the  inclined  web 
member  makes  with  the  vertical.  Using  the  stress  in  X-i  already 
found,  and  resolving  the  forces  at  U1?  the  stress  in  X-2  =  —  JP 
tan  0,  and  the  stress  in  1-2  =  +  -J-  P  sec  0.  Using  the  stresses  in 
Y-i  and  1-2  already  found,  and  resolving  the  forces  at  L±,  the 
stress  in  2-3  = — J-  P  sec  0,  and  the  stress  in  ¥-3  =  -j-  f  P  tan  0. 
In  like  manner,  the  stresses  in  the  remaining  members  may  be 
determined.  Referring  terthe  stresses  found  above,  it  is  seen  that 
all  the  chord  stresses  have  the  common  factor  P  tan  0,  and  that 
all  the  web  members  have  the  common  factor  P  sec  0.  It  is 
further  seen  that  these  factors  are  multiplied  by  coefficients  which 
are  expressed  in  terms  of  the  number  of  panels  in  the  truss. 

These  coefficients  may  be  readily  found  in  the  following  man- 
ner :  In  this  discussion,  it  should  be  borne  in  mind  that  the  chord 
coefficients  are  to  be  multiplied  by  P  tan  0,  and  the  wreb  coefficients 
by  P  sec  0  to  get  the  stresses  in  the  members.  Considering  the 
forces  at  L0,  it  is  seen  that  Rt  =  -J-  and  acts  upward,  therefore,  for 
equilibrium,  X-i  =  -J-  and  acts  downward;  as  indicated  by  the 
arrow.  Since  X-i  =  -J-  and  acts  toward  the  left,  Y-i  =  -J-  and 
acts  toward  the  right.  It  is  seen  from  the  arrows  that  X-i  is 
compression  ( — ),  and  that  Y-i  is  tension  (  +  ). 

Now  consider  the  forces  at  U^  It  has  already  been  shown 
that  X-i  is  compression,  therefore  the  arrow  on  X-i  will  act 
toward  the  joint.  Since  X-i  =  1.  and  acts  upward,  for  equilib- 
rium, 1-2=  -J-  and  acts  downward.  Likewise,  since  both  X-i  and 
1-2  act  toward  the  right,  for  equilibrium,  X-2  =  -J-  +  -J-  =  | -and 
acts  toward  the  left. 

Next  consider  the  forces  at  Lx.  Since  1-2=  1.  and  acts 
upward,  2-3  =  -J  and  acts  downward.  Likewise,  since  Y-i,  1-2, 
and  2-3  all  act  toward  the  left,  for  equilibrium,  Y-3  =  -J-  -f  J-  -f 
-J-  ==  f  and  acts  toward  the  right. 


Art.  7. 


METHOD   OF   COEFFICIENTS. 


261 


In  like  manner,  the  coefficient  and  the  kind  of  stress  may  be 
found  for  each  member  (see  Fig.  135). 

The  chord  stresses  may  be  obtained  by  multiplying  the  coeffi- 
cients by  P  tan  9,  and  the  web  stresses,  by  multiplying  the  coeffi- 
cients by  P  sec  6. 

The  coefficients  for  any  loading  may  be  found  in  the  manner 
indicated  above. 

203.  Loading  for  Maximum  and  Minimum  Live  Load 
Stresses.  The  coefficients  for  all  the  members  for  a  load 
applied  at  each  lower  chord  joint  in  turn  are  shown  in  Fig.  136. 
The  coefficients  shown  in  the  top  line  are  those  for  a  load  on  the 
left  of  the  truss.  The  stresses  in  the  chord  and  web  members  may 
be  obtained  by  multiplying  the  coefficients  by  P  tan  6  and  P  sec  0, 
respectively. 


-  10 
-   8 
-  6 
-  4 
-  2 

,   -? 

-   6 
-  16 
-  12 
-  6 
-  4 
_48 
6 

-   fe 
-  IZ 

-18 
-  IZ 
-   6 

-54 
6 

-  4 
-  8 
-IZ 
-  16 

_48 

-2 

-  4 
-  6 
-  8 

3 

6 

5  • 
4 

L°+"4 

M,! 

11.2+    7 

!Ln,5o 

IE  +  3 

|L',  J  i    Lo 

5 
4 

3 

R,      f   3 

P      +9 

P       +'l5 

P       +  15 

P       +9 

P        +  3     o 

3 

Z 

+  6 

+  10 

+  14 

+  IZ 

+  4     Kz 

Z 

I 

+   I 

+  3 

+    5 

+    7 

+  9 

+  J. 

15 
6 

*£ 

+  ? 

+  5<B 

+  !' 

+  39 
fe 

+  'l 

"? 

Chord   Stresses  =  Coefficients  x  P  Tan  6 

Web    Stresses  =  Coefficients  X  P  sec  6 

Coefficients  for  load  at  each   panel   point 

FIG.  136.     LIVE  LOAD  COEFFICIENTS — WARREN  TRUSS. 

The  stress  in  any  number  due  to  any  loading  may  be  readily 
found  from  Fig.  136.  It  is  seen  that  each  load  produces  compres- 
sion in  the  upper  chord  and  tension  in  the  lower  chord.  The  max- 
imum stresses  in  the  upper  and  lower  chords  are  therefore 
obtained  when  the  truss  is  fully  loaded.  Referring  to  Fig.  136, 
it  is  seen  that  the  maximum  live  load  stresses  in  both  X-i  and  1-2 


262  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXL 

are  obtained  when  the  truss  is  fully  loaded ;  since  each  load  pro- 
duces compression  in  X-i  and  tension  in  1-2.  The  minimum  live 
load  stresses  are  zero,  when  there  are  no  loads  on  the  truss.  The 
maximum  live  load  stresses  in  2-3  and  3-4  are  — l$-  P  sec  0  and 
-f-  V  P  sec  6,  respectively,  when  the  loads  at  L2,  L3,  L2',  and  L/ 
are  on  the  truss.  The  minimum  live  load  stresses  in  2-3  and  3-4 
are  +  -^  P  sec  6  and  —  -JP  sec  0,  respectively,  when  the  load  at  L± 
is  on  the  truss.  When  the  truss  is  fully  loaded,  the  stress  in  2-3 
is  ( —  y  +  i)  P  sec  0  =  — -f-P  sec  0,  and  the  stress  in  3-4  is 
(+ -V1  —  i  )  P  sec  0  —  +  I-P  sec  #•  The  maximum  stresses  in 
4-5  and  5-6  are  —  f  P  sec  0  and  +  -J-  P  sec  0,  respectively,  when 
,  the  loads  at  L3,  L/  and  L/  are  on  the  truss.  The  minimum 
^stresses  in  4-5  and  5-6  are  +  f  P  sec  #  and  —  f  P  sec  6,  respect- 
ively, when  the  loads  at  Lx  and  L2  are  on  the  truss.  When  the 
truss  is  fully  loaded,  the  stress  in  4-5  is  ( —  |- -f  f  )  P  sec  0  = 
—  £  P  sec  0,  and  the  stress  in  5-6  is  ( +  £  —  £  )  P  sec  6  =  -f  f 
P  sec  6. 

Conclusions.  The  following  conclusions  may  be  drawn  from 
the  live  load  coefficients  shown  in  Fig.  136. 

(1)  For  maximum  live  load  chord  stresses,  load  the  truss 
fully  ivith  live  load.    For  minimum  live  load  chord  stresses,  there 
should  be  no  live  load  on  the  truss. 

(2)  For  maximum,  live  load  iveb  stresses,  load  the  longer 
segment  of  the  truss  only. 

(3)  For  minimum  live  load  zveb  stresses,  load  the  shorter 
segment  of  the  truss  only. 

Simplified  Method.  Instead  of  considering  the  coefficients  for 
the  member  due  to  each  separate  load,  it  is  usually  more  convenient 
to  find  the  coefficients  directly  for  all  the  loads.  The  dead  load 
coefficients  for  the  truss  may  be  readily  found  by  first  finding  the 
effective  reactions  in  terms  of  the  panel  load,  and  then  determining 
the  coefficients  from  the  end  toward  the  center  of  the  truss  (see 
Fig.  137,  a).  The  live  load  coefficients  for  the  chords  are  the 
same  as  those  for  the  dead  load.  The  live  load  coefficient  for  the 
maximum  or  minimum  stress  in  any  web  member  may  be  readily 
found  from  the  formula : 


Art.  7. 


METHOD    OF    COEFFICIENTS. 


263 


111 


Live  load  web  coefficient  =• 


Ill 


-,  where 


rn  =  number  of  loads  on  truss  for  a  maximum  or  minimum 

stress  in  any  web  member,  and 
n  =  number  of  panels  in  truss. 

-5  -3  -9  -8  -5 


w(a)'z    w    •? 
Chord  Stresses  =  Coefficients   x  W  tan  9 
Web    Stresses  =  Coefficients  x  W  sec  9 


DEAD 

-8 


LOAD    COEFFICIENTS 

-9          -8 


-5 


I 

P 

Chord  Stresses  =  Coefficients   X   P  tan  9 
Web   Stresses  =  Coefficients  x  P  sec  9 

LIVE   LOAD    COEFFICIENTS 


•«  v* 

Coefficients 


D.L  -  6300 

L.L  -20000 

Max--  26300 
U.    Mirv-   6500 


D.L  - 10000 
L.L. -3zooo 

Max--42000 
Uz   Min-- 10000 


D-L    -12500 

L-L    -40000 

,    Max-- 52500 

Us  Min- -12500 


Lo  D-L  +  3100 
L.L  +10000 
Max-  +  13IOO 
Min-  +  3100 


Li  D-L  +  8100 
L-L  +26000 
Max+34100 
Min-  +  8100 


(O 


D-L  +  10600 
L-L  +34000 
Max- +44600 
Min  + 10600 


Lenqth=  60  ft-,  Heiqht  =  10  ft-,  Width  =  16  ft. 
W=  2500  Ibs-, P=8000  lbs-,Tan  9=0-500;  5ec  6  - 1-116 
FIG.  137.     MAXIMUM  AND  MINIMUM  STRESSES — WARREN  TRUSS. 


264  STRESSES    IN    BRIDGE    TRUSSES.  Chap.  XXI. 

The  second  differences  of  the  maximum  coefficients  in  the  web 
members  are  constant,  which  furnishes  a  check  on  the  work.  The 
numerators  of  these  coefficients  are  I,  3,  6,  10,  etc.,  while  the 
denominator  of  each  is  the  number  of  panels  (see  Fig.  136). 

204.  Coefficients  and  Stresses  in  a  Warren  Truss.     It  is 
required  to  find  the  maximum  and  minimum  stresses  in  all  the 
members  of  the  Warren  truss  shown  in  Fig.  137.    The  truss  has 
a  span  of  60  ft. ;   a  height  of  10  ft. ;   and  a  panel  length  of  10  ft. 
The   dead  panel  load  is   2500  Ibs.,  and  the  live  panel  load   is 
8000  Ibs. 

The  dead  load  coefficients  are  shown  in  Fig.  137,  a.  The  dead 
load  chord  stresses  are  obtained  by  multiplying  the  chord  coeffi- 
cients by  W  tan  0,  and  the  dead  load  web  stresses,  by  multiplying 
the  web  coefficients  by  W  sec  0.  These  stresses  are  shown  on 
the  truss  diagram  in  Fig.  137,  c. 

The  live  load  coefficients  are  shown  in  Fig.  137,  b.  The  chord 
and  web  stresses  are  obtained  by  multiplying  the  coefficients  by 
P  tan  0  and  P  sec  6,  respectively.  These  stresses  are  shown  in 
Fig.  137,  c.  Instead  of  placing  both  the  maximum  and  minimum 
coefficients  on  the  same  member,  the  maximum  coefficients  are 
placed  on  the  member  and  the  minimum  coefficients  on  the  corre- 
sponding member  on  the  other  side  of  the  center  line. 

The  maximum  and  minimum  stresses  are  obtained  by  combin- 
ing the  corresponding  dead  and  live  load  stresses.  These  stresses 
are  shown  in  Fig.  137,  c. 

205.  Coefficients  for  a  Pratt  Truss.     The  dead  load  co- 
efficients for  a  Pratt  truss  are  shown  in  Fig.  138,  a.     The  hip 
vertical  acts  as  a  hanger  to  support  a  single  panel  load,  and  its 
coefficient  is  unity.     The  dead  load  chord  and  web  stresses  may 
be  obtained  by  multiplying  the  coefficients  by  W  tan  0  and  W 
sec  6,  respectively. 

The  live  load  coefficients  are  shown  in  Fig.  138,  b.  The  coeffi- 
cients for  maximum  live  load  stresses  are  shown  on  the  left,  and 
those  for  minimum  stresses,  on  the  right  of  the  center  line.  The 
live  load  chord  and  web  stresses  may  be  obtained  by  multiplying 
the  coefficients  by  P  tan  0  and  P  sec  0,  respectively. 

206.  Coefficients  for  a  Baltimore  Truss.     The  dead  load 


Art.  7. 


METHOD    OF    COEFFICIENTS. 


265 


coefficients  for  a  through  Baltimore  truss  are  shown  in  Fig.  139,  a. 
The  signs  placed  before  the  coefficients  and  the  arrows  on  the 
members  indicate  the  kind  of  stress.  The  sub-members  in  this 
truss  carry  the  loads  to  the  main  members  and  are  not  a  part  of 
the  main  truss  system.  Care  must  be  used  in  determining  the 
coefficients,  as  some  of  the  sub-members  carry  the  loads  toward 
the  center  of  the  truss,  and  the  loads  are  then  carried  back  to  the 
abutment  by  the  main  members.  The  dead  load  chord  and  web 
stresses  may  be  obtained  by  multiplying  the  coefficients  by  W  tan  6 
and  W  sec  0,  respectively. 

-5  -6  -6  -6  -5 


\o  o/ 

\    / 

x 


~ 
W 


+5 


nr 
W 


+3 


W+6 
(a) 

Chord    Stresses  =  Coef f ic lents  x  W  tan  9 
Web    Stresses  =  Coefficients  xWsec6 

DEAD    LOAD  COEFFICIENTS 
-6  -6 


(b)  P  P  ' 

Chord  Stresses  =  Coefficients  x  P  tan  0 
Web    Stresses  =  Coefficients  x  P  sec 6 

LIVE  LOAD   COEFFICIENTS 
FIG.  138.     DEAD  AND  LIVE  LOAD  COEFFICIENTS — PRATT  TRUSS. 

The  live  load  coefficients  are  shown  in  Fig.  139,  b.  The  live 
load  chord  coefficients  are  the  same  as  those  for  the  dead  load. 
The  coefficients  for  maximum  live  load  stresses  are  shown  on  the 
left,  and  those  for  minimum  live  load  stresses,  on  the  right  of 
the  truss.  The  maximum  live  load  stresses  in  the  diagonal  web 
members  whose  coefficients  are  -f-  ft  and  +  H-  are  both  obtained 
when  the  loads  extend  to  the  right  of  the  member  whose  coeffi- 
cient is  +-J-J-.  Likewise,  the  maximum  live  load  stresses  in  the 


266 


STRESSES    IN    BRIDGE    TRUSSES. 


Chap.  XXI. 


members  whose  coefficients  are  —  |-f,  -f  -ff,  and  -j-  ff  are  all 
obtained  when  the  loads  extend  to  the  right  of  the  member  whose 
coefficient  is  -4-  -f-f .  The  former  loading  in  this  particular  case 
will  give  the  same  coefficient.  The  apparent  peculiarity  of  the 

-10* 
Jt^~ 
-6, 


J+6i  Jt6   J+6    JtIO  1+10  J-HZ  J+IZ  J+IO  J-HOj+6  J+6 
W       W       W       W        W       W(a)W        W       W       W       W        W 

Chord  Stresses  =  Coefficients  X  W  tan  9 

Web  Stresses  =  Coefficients  X  W  sec  9 

DEAD    LOAD   COEFFICIENTS 

-Wi 


W    Rt 


|+6i  (  +  6    J  +  6  J  +  IO  J  +  IO  {-HZ 
R,    P        P        P        P       P       P(b)  P       P       P        P       P 

Max>   Chord  Stresses  =  Coefficients  X   P  tan  6 
Web  Stresses  =  Coefficients  X  P  5ec  0 

LIVE    LOAD  COEFFICIENTS 
FIG.  139.     DEAD  AND  LIVE  LOAD  COEFFICIENTS — BALTIMORE  TRUSS. 

loading  for  maximum  stresses  is  due  to  the  sub-members.  Care 
should  be  used  in  getting  the  coefficients  for  minimum  web 
stresses  in  this  type  of  truss.  The  live  load  chord  and  web  stresses 
may  be  obtained  by  multiplying  the  coefficients  by  P  tan  6  and 
P  sec  6,  respectively.  In  making  the  combinations  for  maximum 
and  minimum  dead  and  live  load  web  stresses,  care  must  be  used 
in  determining  whether  or  not  the  counters  are  acting. 


CHAPTER  XXII. 


INFLUENCE    DIAGBAMS,    AND    POSITIONS    OF    ENGINE    AND 
TEAIN   LOADS   FOE   MAXIMUM   MOMENTS,   SHEAES, 
AND    STEESSES. 


This  chapter  will  treat  of  the  construction  of  influence  dia- 
grams, and  of  their  use  in  determining  the  positions  of  wheel 
loads  for  maximum  moments,  shears,  and  stresses  in  girders  and 
trusses.  When  a  series  of  concentrated  loads  moves  across  a 
girder-bridge,  the  maximum  moments  and  shears  at  various 
points  are  usually  given  by  different  positions  of  the  wheel  loads, 
and  in  many  cases  by  a  different  wheel  load  at  each  point.  Like- 
wise in  the  case  of  trusses,  the  maximum  moments  in  chord 
members  and  maximum  shears  in  web  members  are  usually  given 
by  different  positions  of  the  loads,  and  often  by  a  different  wheel 
load  at  the  various  panel  points.  It  is  possible  to  derive  criteria 
for  the  positions  of  wheel  loads  for  maximum  moments  and 
shears  in  girders  and  trusses,  and  this  chapter  will  take  up  the 
determination  of  such  criteria  by  means  of  influence  diagrams. 

207.  Influence  Diagrams.*  Definitions.  An  influence  dia- 
gram is  a  diagram  which  shows  the  variation  of  the  effect  at  any 
particular  point,  or  in  any  particular  member,  of  a  system  of 
loads  moving  over  the  structure.  Influence  diagrams  representing 
the  variations  of  bending  moments  and  shears  in  trusses  and 
beams  are  commonly  used,  and  will  be  taken  up  in  this  chapter. 
The  difference  between  an  influence  diagram  and  a  bending  mo- 
ment or  shear  diagram  is  that  the  former  shows  the  variation  of 
bending  moments  or  shears  at  a  particular  point,  or  in  a  particu- 

*For  a  full  discussion  of  influence  diagrams,  see  a  paper  by  G.  F.  Swain, 
Trans.  Am.  Soc.  C.  E.,  July,  1887. 

267 


268         INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.  Chap.  XXII. 

lar  member,  for  a  system  of  moving  loads ;  while  the  latter  shows 
the  bending  moments  or  shears  at  different  points  for  a  system  of 
fixed  loads. 

The  influence  diagram  is  usually  drawn  for  a  load  unity,  and 
in  this  chapter  only  unit  influence  diagrams  will  be  considered. 
The  moments,  shears,  or  stresses  for  any  system  of  moving  loads 
may  be  obtained  from  the  intercepts  in  the  unit  influence  diagram 
by  multiplying  them  by  the  given  loads. 

The  equation  representing  the  influence  diagram  at  any  point 
may  be  derived  by  writing  the  equation  for  the  function  due  to  a 
load  unity  at  the  point. 

The  principal  use  of  influence  diagrams  is  to  find  the  position 
of  a  system  of  moving  loads  which  will  give  maximum  moments, 
shears,  or  stresses ;  although  they  may  also  be  used  to  determine 
the  values  of  these  functions. 

208.  Position  of  Loads  for  a  Maximum  Moment  at  Any 
Point  in  a  Beam,  or  at  Any  Joint  of  the  Loaded  Chord  of  a 
Truss  with  Parallel  or  Inclined  Chords  (a)  Concentrated 


a  ^dx    b  dx  c 

FIG.  140.     INFLUENCE  DIAGRAM — MOMENT  AT  LOADED  CHORD  JOINT  b'. 

Moving  Loads.     Let  S  Pj  (Fig.  140,  a  and  Fig.  140,  b)  be  the 
summation  of  the  moving  loads  to  the  left  of  the  point  b'  of  the 


MAX.     MOMENT  -  LOADED    CHORD    JOIXT.  269 

truss  or  beam,  and  let  2  P2  be  the  summation  of  the  moving  loads 
to  the  right  of  b'.  It  is  required  to  draw  the  influence  diagram 
for  the  bending  moment  at  b'  ;  also  to  determine  the  position  of 
the  moving  loads  for  a  maximum  bending  moment  at  b'4 

To  construct  the  influence  diagram  (Fig.  140,  c)  for  the 
bending  moment  at  b',  compute  the  moment  at  b'  for  a  load  unity 

d  (L  —  d) 
at  that  point.    This  moment  =  -  1  -  =  ordinate  be.    Draw 

the  horizontal  line  ac,  lay  off  the  ordinate  be,  and  draw  the  lines 
ae  and  ce.  Now  when  the  load  unity  is  to  the  left  of  b',  the  bend- 
ing moments  at  V  are  represented  by  the  ordinates  to  the  line  ae  ; 
and  when  the  load  unity  is  to  the  right  of  b',  by  the  ordinates  to 
the  line  ce. 

A  convenient  method  for  drawing  the  influence  diagram  with- 
out actually  computing  the  moment  will  now  be  shown.  Let  y 
represent  the  ordinate  to  the  line  ae,  and  x,  the  distance  from  the 
load  unity  to  the  left  end  of  the  span.  Then  the  equation  of  the 
line  ae  (i.  e.,  for  the  load  unity  to  the  left  of  b')  is 

(L  —  d)  x(L  —  d) 


and  the  equation  of  the  line  ce  (i.  e.,  for  the  load  unity  to  the 
right  of  b')  is 

d  (L  —  x) 

r-^TT^  <*> 

When  x  =  d,  these  two  equations  have  a  common  ordinate  = 

d  (L  —  d) 

-  =  -  .     When  x  =  L,  the  ordinate  to  the  line  ae  is  L  —  d; 

.Lrf 

and  when  x  =  o,  the  ordinate  to  the  line  ce  is  d.  Therefore,  to 
construct  the  influence  diagram  (Fig.  140,  c),  lay  off  at  a  the 
distance  am  =  d,  and  at  c,  lay  off  en  =  L  —  d.  The  intersection 
of  an  and  cm  will  then  locate  the  point  e  of  the  influence  diagram. 


JThe  criterion  determined  in  this  section  applies  to  the  unloaded  chord 
joints  if  they  are  on  the  same  vertical  lines  as  those  in  the  loaded  chords. 


270         INFLUENCE    DIAGRAMS  —  POSITION    OF    LOADS.  Chap.  XXII. 

The  moment  at  any  point  along  the  truss  or  beam  due  to  any 
number  of  moving  loads  may  be  found  from  the  unit  influence 
diagram  by  multiplying  the  ordinate  under  each  load  by  the  load 
and  taking  the  sum  of  these  products. 

The  position  of  the  loads  for  a  maximum  bending  moment  at 
any  point  b'  will  now  be  determined.  Since  2  Px  and  2  P2  repre- 
sent the  summation  of  all  the  loads  to  the  left  and  to  the  right  of 
b'  ',  respectively,  they  will  have  the  same  effect  as  the  separate 
loads.  The  bending  moment  at  b'  due  to  the  loads  2  Pt  and 
2P2is 

M  =  SPiyi  +  SP2y2.  (3) 

Now  let  the  loads  be  moved  a  small  distance  dx  to  the  left, 
the  movement  being  so  small  that  none  of  the  loads  pass  a',  b', 
or  c'  '.  Then  the  bending  moment  is 

M  +  dM  =  25  P!  (yi  —  dyi)  +  2  P2  (y2  +  <ly2).  (4) 
Subtracting  equation  (3)  from  equation  (4),  we  have 

dM=  —  25  P^  +  S  P2dy2.  (5) 

Referring  to   Fig.    140,   c,  it  is  seen  that  dyi  =  dx  tan  oct, 

be       L—  d 

and  that  dy0  =  dx  tan  oc0.     But  tan  oc    ==—---  —  -  -    and  tan 

ab  L 

be        A 
<x2=  7—  ~=~-     Substituting  these  values  of  dyl  and  dy2  in  equa- 

dM 
tion  (5),  dividing  through  by  dx,  and  placing—:  —  -  =  o  for  a  maxi- 

CIX 

mum,  we  have 

—  „ 

=o. 


Solving  equation  (6),  we  have 


S  Pt      S  Px  +  2  P2 
or  r-=-  -.  (7) 


MAX.     MOMENT LOADED    CHORD    JOINT.  271 

Equation  (7)  is  the  criterion  for  the  maximum  bending  mo- 
ment at  b'  in  a  truss  or  beam.  Since  b'  is  any  point  along  the 
truss  or  beam,  this  criterion  expressed  in  words  is — the  maximum 
bending  moment  occurs  when  the  average  load  to  the  left  of  the 
point  is  equal  to  the  average  load  on  the  entire  span. 

For  a  given  system  of  moving  loads,  it  may  happen  that  two 
or  more  positions  of  the  loads  will  satisfy  the  criterion.  In  this 
case,  the  moment  for  each  position  must  be  found.  As  soon  as 
the  position  of  the  loads  has  been  determined,  the  bending  moment 
may  be  found,  either  algebraically  or  graphically,  by  the  methods 
given  for  fixed  loads. 

If  the  truss  has  equal  panels,  the  most  convenient  unit  of 
length  to  use  is  a  panel  length.  For  a  truss  with  unequal  panels, 
or  for  a  beam,  the  common  unit  of  length  is  the  foot. 

(b)  Uniform  Load.  The  moment  at  b'  (Fig.  140)  for  any 
length  of  uniform  load  of  p  pounds  per  linear  foot  is  equal  to 
the  area  of  the  influence  diagram  included  between  the  extreme 
ordinates  of  the  uniform  load  multiplied  by  the  load  per  linear 
foot.  For,  the  bending  moment  due  to  a  length  dx  of  the  uni- 
form load  =  p  y  dx,  which  is  the  area  under  that  load.  If  the 

length  of  the  uniform  load  — x,  the  moment  at  b'  =   I      p  y  dx 

*/  o 
=  p  X  area  of  influence  diagram  under  the  uniform  load.    For  a 

maximum  bending  moment  at  any  point,  it  is  seen  that  the  span 
must  be  entirely  covered  with  the  uniform  load.  The  maximum 
bending  moment  at  b'  =  p  X  area  of  influence  diagram  aec. 

209.  Position  of  Loads  for  a  Maximum  Moment  at  Any 
Joint  of  the  Unloaded  Chord  of  a  Truss  with  Parallel  or 
Inclined  Chords.  Let  2  P  be  the  total  moving  load  on  the 
truss ;  let  S  Pl  be  the  summation  of  the  moving  loads  to  the  left 
of  b'  (Fig.  141,  a)  ;  let  S  P2  be  the  summation  of  the  moving 
loads  in  the  panel  b'c' ;  and  let  2  P3  be  the  summation  of  the 
moving  loads  to  the  right  of  c'  in  any  truss  with  horizontal  or 
inclined  chords  in  which  the  unloaded  chord  joints  are  not  ver- 
tically over  those  of  the  loaded  chord.  Also,  let  r  =  the  horizontal 
distance  from  e'  to  the  next  loaded  chord  joint  toward  the  left 
end  of  the  truss ;  let  s  =  the  horizontal  distance  from  e'  to  the 


272 


INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.        Chap.  XXII. 


left  end  of  the  truss ;  and  let  L  =  the  span  of  the  truss.  It  is 
required  to  draw  the  influence  diagram  for  the  moment  at  e',  and 
to  determine  the  position  of  the  moving  loads  for  a  maximum 
moment  at  e'. 

Now  it  is  seen  that  the  moment  at  e'  for  a  load  unity  moving 
from  b'  to  a'  and  from  f  to  c'  is  the  same  as  for  a  simple  truss. 
The  portions  of  the  influence  diagram  for  these  parts  of  the 
truss  are  ag  and  fh,  respectively,  which  are  portions  of  the 
influence  diagram  aof.  The  load  in  the  panel  b'c'  is  carried  to 
the  panel  points  V  and  c'  by  the  stringer;  and  the  influence  line 
for  this  part  of  the  truss  is  therefore  the  straight  line  gh. 

The  criterion  for  a  maximum  moment  at  e'  may  be  deter- 
mined in  the  following  manner :  Let  the  loads  be  moved  a  small 


L-s 


Influence  Dioqranrv 

for 

h      Moment  at  Joint  e1 
(bi 


a  b          k          c  f 

FIG.  141.     INFLUENCE  DIAGRAM — MOMENT  AT  UNLOADED  CHORD  JOINT  e'. 

distance  dx  toward  the  left  end  of  the  truss  from  the  position 
shown  in  Fig.  141,  a.     The  increase  in  the  moment  at  e'  will  be 

dM  =  2  P3dx  tan  oc  3  —  2  P2dx  tan  oc  2  —  2  Pxdx  tan  oc ,. 
For  a  maximum, 

dM 


dx 


=  2  P3tan  oc3  — 2P2tan  oc2  —  2  Px  tan  oc1  =  o.        (i) 


MAX.    MOMENT    AT    UNLOADED    CHORD    JOINT.  273 

s                          he  —  gb                                    L  —  s 
But  tan    oc3=- ;    tan    oc2=       — ;    and  tan    oc1==  — 

Also, 

he  =  (L  —  s  +  r  —  d)  tan  oc 3,  and  gb  =  (s  —  r)  tan  ex a. 

Substituting  these  values  of  he  and  gb  in  the  expression  for 
tan  oc2)  and  then  substituting  the  values  of  tan  oc3  and  tan  oc15 
we  have 

rL  —  sd 


dL 

Substituting   the   values   of   tan  cc3,   tan  oc2,   and   tan  ocl   in 
equation  (i),  we  have 


Now  substituting  2  P  for  2  Px  +  2  P2  +  2  P3,  and  reducing, 
we  have 


2P 


SP         .-         , 

_=  -  -  -  ,  (3) 

L  S 

which  is  the  criterion  for  a  maximum  moment  at  any  joint  of  the 
unloaded  chord. 


^-        .1-  VJ, 

Referring  to  equation  (2),  it  is  seen  that-; 

L i  S 

must  become  o  by  passing  from  positive  to  negative,  and  that 
this  can  only  occur  when  some  load  passes  c'  or  b'. 

210.     Position   of   Loads   for   a   Maximum   Moment   at   a 
Panel   Point   of   a   Truss   with    Subordinate   Bracing.      It    is 

required    to    determine   the   position    of   the    wheel    loads    for   a 


274 


INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.         Chap.  XXII. 


maximum  moment  at  the  panel  point  g'  of  the  truss  with  subor- 
dinate bracing  shown  in  Fig.  142,  a.  A  truss  with  subordinate 
bracing  is  one  which  has  points  of  support  for  the  floor  system 
between  the  main  panel  points.  It  is  seen  that  g'  is  the  center 
of  moments  for  determining  the  stress  in  the  lower  chord  member 
b'e'.  The  portion  of  the  load  in  the  panel  c'e',  carried  to  c',  is 
on  the  left  of  the  section  p-p ;  and  its  moment  must  therefore 
be  considered  in  getting  the  moment  at  g'. 


\?  h 


Influence  Diagram 

for 

Moment  at  Joint  cf 
(b) 


be 

FIG.  142.     INFLUENCE  DIAGRAM — TRUSS  WITH  SUBORDINATE  BRACING. 

The  influence  diagram  for  the  moment  at  g'   (Fig.   142',  b) 
is  drawn  as  follows:     For  a  load  unity  moving  from  a'  to  b', 

s  (  L  —  s  ) 

= ;    and  ag  is  the 


the  moment  at  g'  increases  from  o  to. 


influence  line  for  the  segment  a'b'.  Likewise,  the  influence 
line  for  a  load  unity  moving  from  f  to  e'  is  fh.  As  the 
load  unity  moves  from  b'  to  c',  the  moment  at  g'  of  the 
forces  to  the  left  of  section  p-p  is  increased  at  the  same  rate 
as  from  a'  to  b' ;  and  gm  is  the  influence  line  for  the  segment 
b'c'.  The  influence  line  for  the  segment  c'e'  is  the  straight 
line  mh. 

To  determine  the  position  of  the  wheel  loads'  for  a  maximum 
moment  at  g',  let  2  Plf  2  P2,  2  P3,  and  2  P  be  the  loads  in  the 


MAX.    MOMENT — SUBORDINATE    BRACING. 


275 


segment  a'b',  in  the  panel  c'e',  in  the  segment  e'f,  and  on  the 
entire  span,  respectively.  Now  if  the  loads  are  moved  a  small 
distance  dx  to  the  left,  the  moment  at  g'  will  be  increased  by 

dM  =  2  P3dx  tan  « 3  -f-  S  P2dx  tan  oc  2  —  %  P:dx  tan  oc  t. 
For  a  maximum  moment, 

dM 
-: —  =  2  P3  tan  oc3  -f-  2  P2  tan  oc2  —  5  Pt  tan  oc1==o.         (i) 

But, 


and  tan  oc    =• 


nr 


L  —  s 
-  -  ; 

d  tan  cc  j  -f-  2d  tan  oc  3       L  -f-  s 


tan  oc  3  =  -_-  ;   tan  oc  a  = 


rh 


L 


Substituting  these  values  in  equation  (i)  and  reducing,  we  have 
2P       2P,—  2P2 


L  s 

which  is  the  criterion  for  a  maximum  moment  at  g'.  This 
criterion  will  be  satisfied  when  some  wheel  load  is  at  the  panel- 
point  c'. 

2ii.  Position  of  Loads  for  a  Maximum  Shear  at  Any 
Point  in  a  Beam,  (a)  Concentrated  Moving  Loads.  It  is 
required  to  find  the  position  of  the  loads  for  a  maximum  shear 
at  any  point  O  (Fig.  143,  a). 


o  o   o 


r  -+---- 

----_  d 


Influence  Diagram 
Shear  atO 
(b) 


PIG.  143.     SHEAR  IN  A  BEAM. 


The  unit  influence  diagram  (Fig.  143,  b)  will  first  be  drawn. 
For  a  load  unity  entering  the  beam  at  N  and  moving  along  the 


276          INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.          Chap.  XXII. 

L  — S 

span,  it  is  seen  that  the  shear  increases  from  zero  at  N  to  +  — = 

Lt 

when  the  load  is  just  to  the  right  of  O.     When  the  load  passes 
O,  the  shear  at  that  point  is  suddenly  decreased  by  the  amount 


to 


of  the  load  unity,  and  is  equal  to  —  r--,  and  then  increases 

zero  at  M.  For  a  system  of  moving  loads,  it  is  seen  that  the 
positive  shear  increases  as  the  loads  move  toward  the  left  until 
a  load  passes  O,  when  it  is  suddenly  decreased  by  the  amount 
of  the  load.  The  shear  therefore  reaches  a  maximum  every  time 
a  load  reaches  O.  Now  it  may  happen  that,  even  though  the 
loads  are  moved  to  the  left  and  a  load  passes  O,  the  reaction 
will  be  increased  sufficiently  to  increase  the  shear  at  O. 

A  method  will  now  be  shown  for  determining  which  of  the 
two  consecutive  loads,  Px  or  P2,  when  placed  just  to  the  right 
of  O,  will  give  a  maximum  shear. 

Let  2P1  =  total  load  on  the  beam  when  Pt  is  at  O.  Now 
move  the  loads  to  the  left  until  P2  is  just  to  the  right  of  O.  If 
no  additional  load  moves  on  the  beam  and  no  load  moves  off, 
the  shear  will  at  first  be  suddenly  decreased  by  Plt  and  then 

2Pxb 
gradually  increased  by  an  amount  equal  to  2  PJb  tan  oc  =  —  -  —  • 


, 

(see  Fig.  143,  b).   The  total  increase  in  shear  will  be  —  :  —  —  —  Px. 

i_> 

.Pi       2pi 

If  this  expression  is  negative,  i.e.,  if-r—  >  -=  —  ,  then  P±  placed 

at  O  will  give  the  maximum  shear  ;   and  if  it  is  positive,  i.  e.,  if 

P!        S  P!  P!         2  PL 

-r-  <  —  ;:  —  ,  then  P2  will  give  the  maximum  shear.     If-:  —  =  -j  —  , 

then  both  wheels  give  equal  shears. 

If  an  additional  load  moves  on  the  beam  and  no  load  moves 
off  when   P2  moves  up   to   O,   the   expression   representing  the 

2  Pab       PnX 
increase  in  shear  will  be  —  j  —  •  -f-  —z  --  Px  ;    where  Pn  is  the 


MAX.  SHEAR  AT  ANY  POINT  IN  A  BEAM.  .      277 

load  which  enters  the  span,  and  x  is  the  distance  from  the  right 
end  of  the  beam  to  the  load  Pn.  If  2  P2  is  the  total  load  on  the 
beam  when  wheel  P2  is  at  O,  then  the  increase  in  shear  will  lay 

2  Pxb  2  P2b 

between — —  — PI  and  — r PI-  When  the  former  expres- 
sion is  negative  and  the  latter  positive,  then  both  positions  of 
the  loads  should  be  tried.  This  condition  will  occur  for  only  a 
short  distance,  to  the  right  of  which  both  expressions  are  nega- 
tive, and  to  the  left,  both  are  positive.  Wheel  i  at  the  point 

P!      _2Po 

will  give  a  maximum  shear  when  —  —  — — ,  and  wheel  2  will 

D      ^         \  ^ 

p      s  P 
.         ,      ri *  ri 

give  a  maximum  shear  when-r— — — — •  . 

D      ^       J  ^ 

(b)  Uniform  Load.  From  Fig.  143,  b,  it  is  seen  that  the 
maximum  shear  at  any  point  O  due  to  a  uniform  load  will  occur 
when  the  load  extends  from  the  right  end  of  the  beam  to  the 
point  O.  The  minimum  shear  will  occur  when  the  load  extends 
from  the  left  end  of  the  beam  to  the  point  O. 

212.  Position  of  Loads  for  a  Maximum  Shear  in  Any 
Panel  of  a  Truss  with  Parallel  or  Inclined  Chords,  (a)  Con- 
centrated Moving  Loads.  Let  Fig.  144,  a  represent  a  truss 
loaded  with  concentrated  loads.  It  is  required  to  determine  the 
position  of  the  loads  for  a  maximum  shear  in  any  panel,  say 
b'c'.  Let  S  P!  represent  the  total  load  on  the  left  of  this  panel, 
2  P,  that  on  the  panel,  and  S  P3  the  total  load  on  the  right  of  the 
panel  b'c'.  Let  m  =  number  of  panels  on  the  left  of  the  panel 
b'c',  and  n  =  total  number  of  panels  in  the  truss. 

The  influence  diagram  (Fig.  144,  b)  is  constructed,  as  fol- 
lows :  For  a  load  unity  moving  from  a'  to  b',  the  shear  in  the 

m 

panel  b'c'  increases  from  zero  when  the  load  is  at  a'  to 

n 

when  the  load  is  at  b' ;  and  ab  is  the  influence  line  for  the  load 
to  the  left  of  the  panel  b'c'.  For  a  load  unity  moving  from  (V 

n  —  m  —  I 
to  c',  the  shear  increases  from  zero  to  +  ;    and  dc 


278          INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.          Chap.  XXII. 


is  the  influence  line  for  the  loads  to  the  right  of  the  panel  b'c'. 

m 
For  a  load  unity  on  the  panel  b'c',  the  shear  varies  from 


for  the  load  at  b'  to  + 


m —  i 


for  the  load  at  c' ;  and  be  is 


the  influence  line  for  the  load  on  this  panel.  The  load  in  the 
panel  b'c'  is  carried  to  the  joints  b'  and  c'  by  the  stringers,  the 
amount  of  load  transferred  to  each  varying  inversely  as  the  dis- 
tance from  the  load  to  the  joint. 

The  influence  diagram  for  the  entire  span  is  abed,  and  it  is 
seen  that  the  lines  ab  and  be  are  parallel,  and  that  the  perpendicu- 
lar distance  between  them  is  unity. 

Referring  to  the  influence  diagram  (Fig.  144,  b),  it  is  seen 
that  for  a  maximum  positive  shear  in  b'c'  due  to  a  moving  load, 


(a)   i 

i     Influence  Diaqram 

i  Shear  in  b'c1 


FIG.  144.     SHEAR  IN  A  TRUSS. 

the  load  should  be  placed  at  c' ;    and  for  a  maximum  negative 
shear,  the  load  should  be  placed  at  b'. 

The  maximum  positive  shear  in  the  panel  b'c'  is 

S  =  2Py-f-5Py  !EPyi.  (i) 

Now  move  the  loads  a  small  distance  dx  to  the  left,  the  dis- 


MAX.    SHEAR   IN   ANY    PANEL   OF   A   TRUSS.  279 

tribution  of  the  loads  remaining  the  same  as  before.    The  shear 
is  then 

S  +  dS  =  2P3  (y3+dy3)  +  2§P2  (y2—  dy2)—  2P,  (YI—  dyx)  (2). 
Subtracting  equation  (i)  from  equation  (2),  we  have 

dS  =  2  P3dy3  —  2  P2dy2  +  5  P±dyi.  (3) 

But  dy3  =  dxtan  oc  3  =  dx  —  =  , 

nd 

n  —  I 

dy2  =  dx  tan  oc  „  =  dx 


, 
nd 

and  dyx  =  dx  tan  cc  x  =  dx  —  -. 

Substituting  these  values  of  dy3,  dy2,  and  dy±  in  equation  (3), 
and  dividing  through  by  dx,  we  have 


ds 
Placing  —  =  o  for  a  maximum,  and  multiplying  through  by 

C1X 

nd,  we  have 

2P3  —  2P2(n  —  i)+SP1  =  o,  (5) 

2  Pi  +  2  P2  +  5  P, 
from  which       2  P2  =  n       -  2  ,  (6) 

which  is  the  criterion  for  the  maximum  shear  in  any  panel.  This 
criterion  expressed  in  words  is  —  the  maximum  shear  in  any 
panel  of  a  truss  will  occur  when  the  load  in  the  panel  is  equal  to 
the  total  load  on  the  truss  divided  by  the  number  of  panels.  This 
is  equivalent  to  saying  that  the  average  load  in  the  panel  must  be 
equal  to  the  average  load  on  the  entire  bridge. 

This  criterion  requires  that  some  wheel  near  the  head  of  the 
train  shall  be  at  the  panel  point  to  the  right  of  the  panel  in 
which  the  shear  is  required.  Any  particular  wheel  P  placed  at 
the  right-hand  panel  point  will  give  a  maximum  shear  in  that 
panel  if  the  entire  load  on  the  bridge  lays  between  S  P2n  and 


280  INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.         Chap.  XXII. 

(2  Po  +  P)  n;  where  2  P2  is  the  load  in  the  panel  other  than  P, 
and  n  is  the  number  of  panels  in  the  truss. 

(b)  Uniform  Load.  Referring  to  Fig.  144,  b,  it  is  seen  that 
the  maximum  shear  in  the  panel  bV  due  to  a  uniform  load  will 
occur  when  the  load  extends  from  the  right  end  of  the  truss  to 
the  point  h,  i.  e.,  to  the  point  where  the  shear  changes  sign.  The 
minimum  shear  will  occur  when  the  uniform  load  extends  from 
the  left  end  of  the  truss  to  the  point  h.  By  minimum  shear  is 
meant  the  greatest  shear  of  an  opposite  kind. 

213.  Position  of  Loads  for  a  Maximum  Stress  in  any  Web 
Member  of  a  Truss  with  Inclined  Chords.  The  criterion  for 
maximum  stresses  in  chord  members  is  the  same  for  trusses 
with  parallel  chords  as  for  those  with  inclined  chords.  The 
criterion  for  maximum  stresses  in  web  members  for  trusses  with 
inclined  chords  differs  from  that  for  those  with  parallel  chords ; 
since  in  the  former  case,  the  inclined  chord  members  take  a  part 
of  the  shear! 

Let  g'c'  (Fig.  145,  a)  be  any  web  member  of  a  truss  with 
inclined  chords ;  let  p-p  be  a  section  cutting  g'h',  g'c',  and  b'c'  ; 
and  let  O  be  the  point  of  intersection  of  g'h'  and  b'c',  which  is 
the  center  of  moments  for  determining  the  stress  in  g'c'.  Also 
let  2  P  be  the  entire  moving  load  on  the  truss  when  there  is  a 
maximum  stress  in  g'c' ;  let  2  P!  be  the  summation  of  the  mov- 
ing loads  in  the  panel  b'c' ;  and  let  2  P2  be  the  summation  of 
the  moving  loads  to  the  right  of  c'.  It  is  required  to  draw  the 
influence  diagram  for  the  moment  at  O,  and  to  determine  the 
position  of  the  moving  loads  for  a  maximum  stress  in  g'c'. 

The  stress  in  g'c'  is  equal  to  the  moment  of  the  external 
forces  to  the  left  of  the  section  p-p  about  the  point  O  divided 
by  the  moment  arm  r;  and  the  stress  in  this  member  is  a 
maximum  when  the  moment  at  O  is  a  maximum. 

L  —  s  —  d 
The  moment  at  O  for  a  load  unity  at  c'— —        — -y—     -  k, 

L  — s 
and  for  a  load  unity  at  b'  =  -| _—  -   k  —  k  —  s^      The  mo- 


MAX.     WEB     STRESS — INCLINED     CHORDS. 


281 


ment  passes  through  zero  when  the  load  is  at  some  point  between 
b'  and  c'.  The  influence  diagram  for  the  moment  at  O  may 
therefore  be  constructed  by  laying  off  ch  (Fig.  145,  b)  = 

L  —  s  —  d  L  —  s 

— k   (downward  from  c),  bg  =  — = k  —  k  —  s  (up- 

1  >  1^ 

ward  from  b),  and  drawing  ag,  gh,  and  hf. 


Influence  Diaqram 

for 
Moment  at  Point  0 

(b) 


FIG.  145.     STRESS  IN  WEB  MEMBER — TRUSS  WITH  INCLINED  CHORDS. 

The  maximum  stress  in  gV  occurs  when  some  of  the  wheels 
near  the  head  of  the  train  are  in  the  panel  b'c'.  It  will  be 
assumed  in  this  case  that  there  are  no  loads  to  the  left  of  b' ; 
as  this  is  the  usual  condition. 

The  criterion   for  a  maximum  stress  in  g'c'  may  be  deter- 


282  INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.        Chap.  XXII. 

mined  in  the  following  manner:  Let  the  loads  be  moved  a 
small  distance  dx  toward  the  left  end  of  the  truss  from  the 
position  shown  in  Fig.  145,  a.  The  increase  in  the  moment  at 
O  will  be 

dM  =  S  P2dx  tan  oc  3  —  %  Pjdx  tan  ex  2. 
For  a  maximum, 

dM 

2P2tan  «,-— SPjtan  ^2=o.  (i) 


dx 


But  tan  oc  3  =  —  _g =  —  —  ,  since  ch  =  —        — j—      -  k  ; 


and  tan  oc ,  = 


ch  k  L  —  s  —  d 

-  s  — 
ch  +  gb 


d 

L  —  s  —  d .  ,        ,  L  —  s 


_  k         k  +  s 

"17"      ~d       ' 

Substituting  these  values  of  tan  oc  3  and  tan  oc  2  in  equation  ( I ) , 
and  putting  2  P  instead  of  3  Pl  +  2  P,,  we  have 


2P 


d 

which  is  the  criterion  for  a  maximum  moment  at  O  and  therefore 
for  a  maximum  stress  in  the  member  g'c'. 

By  comparing  the  above  criterion  with  that  given  for  a 
maximum  shear  in  §  212,  it  is  seen  that  they  are  very  similar. 
the  only  difference  being  that  in  this  case  the  load  in  the  panel 

is  to  be  increased  by  the  ^  th  part  of  itself  before  dividing  by 
the  panel  length. 


MAXIMUM   FLOORBEAM   REACTION. 


283 


To  get  the  maximum  stress  in  the  member  g'b'  (Fig.  145,  a), 
the  same  panel  b'c'  is  partially  loaded  and  k  is  replaced  by  k', 
the  other  quantities  remaining  the  same  as  for  g'c'. 

214.     Position  of  Loads  for  a   Maximum  Floorbeam  Re- 
action.    It  is  required  to  find  the  maximum  load  on  the  floor- 
beam  at  O  (Fig.  146,  a)  due  to  loads  on  the  panels  MO  and  ON 
The  loads  are  carried  to  M,  O,  and  N  by  the  floor  stringers. 


N' 


e  a1  e1 

(a)     Influence   Diagram  for  (b) 
Shear  at  0  Moment  at  0' 

FIG.  146.     FLOORBEAM  REACTION. 

Fig.  146,  a  shows  the  influence  diagram  for  the  floorbeam 
load  at  O.  The  influence  line  for  the  load  in  the  panel  MO 
is  ac  ;  and  that  for  the  load  in  the  panel  ON  is  cb.  The  ordinate 
ce  is  equal  to  unity. 

Let  M'N'  (Fig.  146,  b)  be  a  beam  whose  length  dx  -f-  d.,  is 
equal  to  the  length  of  the  two  panels  MN  of  the  truss,  and  let 
a'b'c'  be  the  influence  diagram  for  the  bending  moment  at  O', 
whose  distance  from  the  left  end  of  the  beam  equals  dx.  The 

djd, 
ordinate  under  O'  equals  -  —  —  r  . 


By  comparing  the  diagrams  in  Fig.  146,  a  and  Fig.  146,  b, 
it  is  seen  that  they  differ  only  in  the  value  of  the  ordinates  ce 
and  c'e'.  It  is  also  seen  that  the  maximum  floorbeam  reaction 
occurs  for  the  same  position  of  the  loads  as  does  the  maximum 
bending  moment.  The  ratio  of  any  two  corresponding  ordinates 


284         INFLUENCE    DIAGRAMS — POSITION    OF    LOADS.  Chap.  XXII. 

d^L,          dx  +  cl, 
yi  and  y2  is  as  ce  is  to  cV    =  i  -^-^  ^g^- 

Therefore,  the  maximum  floorbeam  reaction  may  be  obtained  by 
finding  the  maximum  bending  moment  at  a  distance  dx  from 
the  end  of  a  beam  whose  length  is  equal  to  the  sum  of  the  two 

panel  lengths  dx  +  d2,  and  multiplying  this  moment  by  1 ,  "  • 
If  the  two  panel  lengths  are  equal,  the  maximum  moment  at 
the  center  of  the  beam  should  be  multiplied  by  _  where  d  is 
equal  to  the  panel  length. 


CHAPTER  XXIII. 

MAXIMUM  MOMENTS,  SHEARS,  AND  STRESSES  DUE  TO  ENGINE 
AND   TRAIN   LOADS. 

This  chapter  will  treat  of  the  determination  of  maximum 
moments,  shears,  and  stresses  due  to  engine  and  train  loads. 
Graphic  methods  of  applying  the  criteria  derived  in  Chapter  XXII 
will  also  be  shown.  The  dead  load  moments,  shears,  and  stresses 
will  not  be  considered ;  as  they  may  be  found  by  the  methods 
already  explained. 

The  chapter  will  be  divided  into  two  articles,  as  follows: 
Art.  i,  Maximum  Moments,  Shears,  and  Stresses  in  Any  Par- 
ticular Girder  or  Truss ;  and  Art.  2,  Maximum  Moments,  Shears, 
and  Stresses  in  Girders  and  Trusses  of  Various  Types  and  Spans. 


ART.  i.      MAXIMUM  MOMENTS,  SHEARS,  AND  STRESSES  IN  ANY 
PARTICULAR  GIRDER  OR  TRUSS. 

When  it  is  required  to  determine  the  maximum  moments, 
shears,  and  stresses  in  any  particular  span,  the  methods  which 
will  now  be  given  will  be  found  convenient.  These  methods  will 
be  explained  by  the  solution  of  two  problems,  involving  (i)  the 
plate  girder,  and  (2)  the  Pratt  truss. 

215.  (i)  Maximum  Flange  Stresses  and  Shears  in  a 
Plate  Girder.  Let  Fig.  147,  a  represent  one  girder  of  a  single- 
track  deck  plate  girder-bridge  whose  span,  center  to  center  of  end 
bearings,  is  60  ft.,  and  whose  effective  depth  is  6  ft.  It  is  required 
to  find  the  maximum  flange  stresses  and  shears  at  the  tenth  points 

285 


286 


MAX.  MOMENTS  AND  SHEARS — ENGINE  LOADS.      Chap.  XXIII. 


along  the  girder  due  to  the  engine  and  train  loading  shown  in 
Fig.  147,  c,  the  loading  being  that  for  one  girder. 

Flange  Stresses.     The  diagram  for  flange  stresses  is  shown 
in  Fig.  147,  b.    To  construct  this  diagram,  lay  off  the  engine  dia- 


O  \ 


3  4  5  4'  3'          2' 


.  ]0_spqces  .flL&QI  r.  BOrQ.  _ 


C 

X] 

L' 

(d 
She 

) 
ars 

> 

^ 

£ 

-  - 

!o 

r  1 

So 

n     f  i 

c 

si 

V          ' 

nr 


3'  V 

0'        5'        10' 


ZO1 


5          4'          3'          Z 

FLANGE  STRESSES  AND  SHEARS 

IN  A 

PLATE  GIRDER  UNDER  ENGINE  AND  TRAIN  LOAD 
5pan=60'-0";  Depth  Ceff)  =  6-0";  Loading  as  5hown. 
All  loads  and  stresses  are  in  "thousands  of  pounds- 


0' 


Scale  of  Distances 

200  400 


Scale  of  Flange  Stresses 
0  50  100 

Stale  of   Loads  and   Shears 


FIG.  147.     MAXIMUM  FLANGE  STRESSES  AND  SHEARS  IN  A  PLATE  GIRDER. 


Art.  1.  PLATE  GIRDER — -MAXIMUM   FLANGE  STRESSES.  287 

gram  (Fig.  147,  c)  to  any  convenient  scale,  and  mark  the  begin- 
ning, middle  point,  and  end  of  the  uniform  load.  Since  the  span 
of  the  girder  is  60  ft.,  it  will  be  necessary  to  consider  only  about 
20  ft.  of  uniform  load.  Lay  off  the  load  line  (Fig.  147,  e)  for 
the  given  engine  and  20  ft.  of  train  load,  and  with  a  pole  distance 
equal  to  some  multiple  of  the  effective  depth  of  the  girder  (in  this 
case  four  times  the  depth),  construct  the  funicular  polygon 
ABGKO  (Fig.  147,  b).  The  portion  of  the  funicular  polygon 
KO  for  the  uniform  train  load  is  an  arc  of  a  parabola,  tangent  to 
KL  at  K  and  to  LO  at  O,  the  method  used  in  constructing  the 
parabola  being  that  shown  in  Fig.  148,  b.  Prolong  the  line  AB 
as  far  as  necessary  to  complete  the  construction.  Now  divide 
the  span  of  the  girder  into  ten  equal  parts,  and  drop  verticals 
from  these  points  of  division  to  meet  the  funicular  polygon. 
The  line  b'b'  (Fig.  147,  b),  which  connects  the  points  of  inter- 
section of  verticals  60  feet  apart  with  the  funicular  polygon, 
is  the  closing  string  of  the  funicular  polygon  for  the  loads  that 
come  upon  the  girder  when  it  is  in  the  position  shown  in  Fig. 
147,  a.  For  this  position  of  the  girder,  wheel  2  is  at  section  2. 
The  ordinates  under  the  several  points  of  division,  intercepted 
between  the  closing  string  b'b'  and  the  funicular  polygon,  repre- 
sent to  scale  the  flange  stresses  at  the  several  points  along  the 
girder  for  this  position  of  the  load.  These  ordinates  represent 
actual  flange  stresses,  and  the  scale  used  in  measuring  them  is 
four  times  that  used  for  the  loads  in  the  load  line ;  since  the  pole 
distance  was  taken  equal  to  four  times  the  depth  of  the  truss 
(see  §  184).  Now  consider  the  girder  to  be  shifted  one  division, 
or  6  feet,  toward  the  right  from  the  position  shown  in  Fig.  147,  a. 
(This  has  the  same  effect  as  moving  the  loads  6  feet  to  the  left.) 
The  closing  string  of  the  funicular  polygon  for  this  new  position 
of  the  loads  and  truss  is  cV.  The  ordinate  directly  above  wheel 
2,  intercepted  between  the  closing  string  c'c'  and  the  funicular 
polygon,  represents  the  flange  stress  at  section  I  of  the  girder; 
and  the  other  ordinates  represent  the  flange  stresses  at  various 
points  along  the  girder.  Likewise,  d'd'  and  e'e'  are  closing 
strings  for  the  girder  moved  two  and  three  divisions  to  the  right, 
respectively,  from  its  original  position;  and  the  flange  stresses 


288  MAX.  MOMENTS  AND  SHEARS — ENGINE  LOADS.     Chap.  XXIII. 

at  different  points  along  the  girder  are  represented  by  the  ordi- 
nates  between  these  closing  strings  and  the  funicular  polygon. 
Also,  a'a'  is  the  closing  string  for  the  funicular  polygon  when 
the  girder  is  moved  one  division  to  the  left  from  the  position 
shown  in  Fig.  147,  a.  From  an  inspection  of  the  diagram  shown 
in  Fig.  147,  b,  it  is  seen  that  it  is  unnecessary  to  draw  any  other 
closing  strings  for  this  problem.  Now  if  the  curve  MM  is  drawn 
through  the  points  distant  each  one  division  horizontally  from 
b',  c',  d',  and  e',  respectively,  it  is  seen  that  the  ordinates  between 
this  curve  and  the  funicular  polygon  will  represent  the  successive 
flange  stresses  at  section  i  as  the  girder  is  moved  to  the  right, 
i.  e.,  when  the  load  moves  along  the  girder  toward  the  left.  Like- 
wise, the  curve  NN,  drawn  through  the  points  distant  each  two 
divisions  horizontally  from  a',  b',  c',  d',  and  e',  respectively,  will 
represent  the  successive  flange  stresses  at  section  2  as  the  load 
moves  along  the  girder.  Also,  the  curves  RR,  SS,  and  TT  repre- 
sent the  successive  flange  stresses  at  sections  3,  4,  and  5,  respect- 
ively, as  the  load  moves  along  the  girder.  By  scaling  the  ordi- 
nates at  the  different  points,  it  is  seen  that  the  maximum  flange 
stress  at  section  i  occurs  when  wheel  2  is  at  section  i,  and  is 
82  ooo  Ibs.  Likewise,  it  is  seen  that  the  maximum  flange  stress 
at  section  2  occurs  when  wheel  3  is  at  section  2,  and  is  143  ooo 
Ibs. ;  that  the  maximum  flange  stress  at  section  3  occurs  when 
wheel  3  is  at  section  3,  and  is  186000  Ibs.;  that  the  maximum 
flange  stress  at  section  4  occurs  when  wheel  4  is  at  section  4,  and 
is  210000  Ibs.;  and  that  the  maximum  flange  stress  at  section  5 
occurs  when  wheel  5  is  at  section  5,  and  is  212  ooo  Ibs.  It  should 
be  noted  that  the  maximum  moment  and  flange  stress  at  any 
point  always  occurs  when  some  wheel  is  at  the  point,  i.  e.,  the 
maximum  ordinate  is  always  at  some  vertex  of  the  funicular  poly- 
gon. If  the  ordinate  is  not  at  one  of  the  division  points  through 
which  the  curves  MM,  NN,  etc.,  were  drawn,  its  length  may  be 
more  accurately  determined  by  drawing  the  closing  line  for  the 
required  position  of  the  loads.  It  is  seen  that  the  construction 
shown  in  Fig.  147,  b  not  only  gives  the  maximum  stresses,  but 
also  the  wheels  which  cause  these  stresses.  Referring  to  §  159, 
it  is  seen  that  the  maximum  moment,  and  therefore  the  maximum 


Art.  1.  PLATE    GIRDER MAXIMUM     SHEARS.  289 

flange  stress  in  the  girder,  occurs  at  some  wheel  near  the  center 
of  the  span  when  that  wheel  is  as  far  to  one  side  of  the  center 
as  the  center  of  gravity  of  all  the  loads  is  to  the  other  side  of  the 
center.  The  position  of  the  center  of  gravity  of  all  the  wheels  on 
the  girder  at  this  time  is  given  by  producing  the  extreme  strings 
AB  and  KL  (Fig.  147,  b)  until  they  intersect.  It  is  seen  that  the 
center  of  gravity  of  all  the  wheels  is  0.4  ft.  to  the  right  of  wheel 
5 ;  therefore  the  maximum  flange  stress  in  the  girder  occurs  when 
wheel  5  is  0.2  ft.  to  the  left  of  the  center.  A  part  of  the  closing 
string  for  this  position  of  the  wheels  is  shown  by  the  dotted  line 
near  b'  (Fig.  147,  b)  ;  and  the  maximum  flange  stress  in  the 
girder  is  213  ooo  Ibs. 

Shears.  The  maximum  live  load  shears  at  the  tenth  points 
may  be  obtained  from  the  diagram  shown  in  Fig.  147,  d.  In  con- 
structing this  diagram,  use  the  same  load  line  as  for  flange 
stresses,  but  take  the  pole  at  P'.  The  pole  P'  is  located  on  a  hori- 
zontal line  through  the  top  of  the  load  line  with  a  pole  distance 
equal  to  the  span  of  the  girder,  60  feet.  Draw  the  funicular  polygon 
A'B'C'G'K'O'  (Fig.  147,  d)  for  the  given  loading,  the  portion 
K'O'  being  the  arc  of  a  parabola  tangent  to  K'L'  at  K'  and  to 
L'O'  at  O'.  With  the  first  driver,  wheel  2,  at  the  left  end  of  the 
girder,  lay  off  the  span  to  the  right,  and  divide  it  into  tenths,  as 
shown  in  Fig.  147,  f.  The  ordinate  dd"  above  the  right  end  of 
the  girder,  intercepted  between  the  horizontal  line  A'B'N'  and 
the  funicular  polygon  A'B'C'G'K'O',  then  represents  the  reac- 
tion at  the. left  end  of  the  span  (see  §  187)  ;  provided  none  of  the 
loads  used  in  the  construction  of  the  funicular  polygon  are  off  of 
the  girder.  Now  when  wheel  2  is  at  the  left  end  of  the  girder, 
wheel  i  is  off  of  the  span,  and  should  not  be  considered  in  draw- 
ing the  funicular  polygon,  i.  e.,  the  horizontal  line  A'B'N'  should 
be  replaced  by  the  line  B'C'rs.  Since  there  are  no  loads  on  the 
girder  to  the  left  of  wheel  2,  the  maximum  shear  at  the  left  end 
of  the  span  is  equal  to  the  left  reaction,  which  is  represented  by 
the  ordinate  sd"  — 98000  Ibs.  When  wheel  2  is  at  section  i, 
which  is  6  feet  from  the  left  end  of  the  girder,  wheel  i  is  still  off 
of  the  girder,  and  the  maximum  shear  at  section  I  is  represented 
by  vc"  =  82  400  Ibs.  The  scale  used  in  measuring  these  ordi- 


290  MAXIMUM    STRESSES — ENGINE    LOADS.        Chap.  XXIII. 

nates  is  the  same  as  that  used  for  the  loads.  Now  if  sections  2,  3, 
and  4  (Fig.  147,  f)  are  successively  placed  under  wheel  2,  then 
the  intercepts  above  the  right  end  of  the  girder  will  represent  the 
left  reactions.  The  shears  at  these  points  are  equal  to  the  reac- 
tions minus  the  weight  of  the  pilot  wheel  I ;  and  are  represented 
by  the  intercepts  between  the  funicular  polygon  and  the  line  mr, 
the  vertical  distance  between  A'B'N'  and  mr  representing  the 
weight  of  wheel  i.  The  shears  at  sections  2,  3,  and  4  are  67  400, 
53  ooo,  and  39  700  Ibs.,  respectively.  When  wheel  2  is  at  section  5, 
the  center  of  the  girder  (see  Fig.  147,  h),  the  shear  at  section  5  is 
27900  Ibs.;  when  wheel  i  is  at  section  5  (Fig.  147,  g),  the  shear 
at  section  5  is  24  500-  pounds ;  and  when  wheel  3  is  at  section  5 
(Fig.  147,  i),  the  shear  at  that  section  is  17800  Ibs.  For  maxi- 
mum shears,  it  is  unnecessary  to  consider  any  points  to  the  right 
of  the  center  of  the  girder.  It  is  seen  that  wheel  2  gives  the 
maximum  shears  at  all  points  from  the  left  end  to  the  center  of 
the  span.  It  may  be  easily  determined  by  trial,  from  the  diagram 
shown  in  Fig.  147,  d,  which  wheel — wheel  I  or  wheel  2 — gives 
the  maximum  shear  at  any  point  along  the  girder,  or  it  may  be 
determined  directly  by  the  criterion  given  in  §  211. 

Another  method  of  obtaining  the  maximum  shears  will  now 
be  given,  in  which  the  shears  are  determined  from  the  diagram 
for  flange  stresses.  When  wheel  2  is  at  the  left  end  of  the 
girder,  wheel  I  is  off  the  span,  the  closing  string  of  the  funicular 
polygon  is  d'd'  (Fig.  147,  b),  and  the  ray,  drawn  through  P 
parallel  to  d'd'  to  intersect  the  load  line,  will  determine  the  two 
reactions  (Fig.  147,  e),  the  left  reaction  being  98000  Ibs.,  which 
is  the  shear  at  the  left  end  of  the  girder.  Likewise,  when  wheel  2 
is  at  section  I,  wheel  I  is  still  off  the  girder,  the  closing  string 
of  the  funicular  polygon  is  cV ;  and  the  ray,  drawn  through  P 
parallel  to  c'c',  will  determine  the  two  reactions,  the  left  reaction 
being  82  400  Ibs.,  which  is  the  shear  at  section  2  of  the  girder, 
Likewise,  when  wheel  2  is  at  section  2,  wheel  i  is  on  the  girder, 
the  closing  string  is  b'b',  the  left  reaction  is  77  400  Ibs.,  and  the 
shear  is  77  400  —  10  ooo  =  67  400  Ibs.  In  a  similar  manner,  the 
shears  at  sections  3,  4,  and  5  of  the  girder  are  found  to  be  53  ooo, 
39700,  and  27900  Ibs.,  respectively.  In  this  problem,  wheel  2  is 


Art.   1.  PRATT  TRUSS — MAXIMUM   CHORD  STRESSES.  291 

at  one  of  the  sections.  If  the  wheel  causing  maximum  shears  is 
not  at  one  of  the  sections,  then  another  set  of  verticals  and  closing 
lines  should  be  drawn  to  determine  the  maximum  shears.  In 
some  cases,  this  method  gives  the  simpler  solution;  while  in 
others,  the  first  method  described  is  more  efficient. 

In  the  problem  given,  the  loading  consisted  of  one  engine  and 
tender  followed  by  a  uniform  train  load.  If  the  loading  consists 
of  two  engines  and  tenders  followed  by  a  uniform  train  load,  the 
maximum  shears  will  be  somewhat  larger.  This  is  evident  from 
the  fact  that  the  straight  line  AB  (Fig.  147,  b)  would  then  be 
replaced  by  a  broken  line,  thus  increasing  the  length  of  the  ordi- 
nates  between  the  closing  lines  and  the  funicular  polygon. 

The  process  explained  is  more  efficient  if  the  equidistant  ver- 
ticals, which  represent  the  divisions  of  the  girder,  are  drawn  upon 
a  separate  sheet  of  tracing  cloth  or  transparent  paper;  as  the 
position  of  the  girder  may  then  be  readily  shifted. 

The  method  explained  in  this  section  affords  an  efficient  solu- 
tion for  any  particular  span  and  loading;  and  if  a  large  scale  is 
used  and  care  is  exercised  in  making  the  constructions  very  good 
results  may  be  obtained. 

216.  (2)  Maximum  Chord  and  Web  Stresses  in  a  Pratt 
Truss.  Let  Fig.  148,  a,  represent  one  truss  of  a  single-track 
through  Pratt  truss-bridge,  whose  span  is  120  ft.,  and  whose 
depth  is  28  ft.  It  is  required  to  find  the  maximum  chord  and 
web  stresses  due  to  the  engine  and  train  loading  shown  in  Fig. 
148,  c,  the  loading  shown  being  that  for  one  truss. 

Chord  Stresses.  In  this  problem,  the  chords  are  parallel,  and 
the  upper  chord  panel  points  are  directly  over  those  of  the  lower 
chord,  which  simplifies  the  solution  somewhat.  The  construc- 
tions and  methods  used  for  this  problem  are  similar  to  those  used 
for  the  plate  girder  (see  Fig.  147,  and  §  215)  ;  and  only  the 
points  in  which  the  methods  differ  for  the  two  cases  need  be 
explained.  The  diagram  for  chord  stresses  is  shown  in  Fig. 
148,  b.  By  an  inspection  of  the  truss  and  loading,  it  was  seen 
that  about  80  feet  of  uniform  train  load  would  be  sufficient.  The 
load  line  is  shown  in  Fig.  148,  f ;  and  since  the  pole  distance  was 
taken  equal  to  twice  the  depth  of  the  truss,  the  ordinates  in  the 


292 


MAXIMUM    STRESSES — EXGIXE    LOADS.         Chap.  XXIII. 


U, 


ut          u; 


Lo  / ..         \|Lt         ML*         MI:*      ^±^ JyVi. 

\ ^pqn.eij.elol-ojji  izPiPl  __ ___  _  -___.__--.-  ^ ,  -  _  r 

^Tc       |d       ft  ,(a)  ,  .a       I         cj. 


T^1       -  L'o 

CHORD  AND  WEB  STRESSES 
PRATT  TRUSS  UNDER  ENGINE  AND  TRAIN  LOAD        Scale  of  Chord  Stresses 

Span  =  120'-0"j  Depth  =  28L0'-  Loading  as  shown  \    ,   .    .    .    i J 

All  loads  and  stresses  are  in  thousands  of  pounds  Scale  of  Web  Stresses 

FIG.  148.     MAXIMUM  CHORD  AND  WEB  STRESSES  IN  A  PRATT  TRUSS. 


Art.  1.  PRATT  TRUSS MAXIMUM  WEB  STRESSES.  293 

chord  stress  diagram  should  be  measured  by  a  scale  equal  to  twice 
that  used  for  the  loads.  The  chord  stresses,  together  with  the 
ordinates  which  represent  them,  are  shown  in  Fig.  148,  b.  From 
an  inspection  of  this  diagram,  it  is  seen  that  the  maximum  stress 
in  L0L2  is  88  ooo  Ibs.  when  wheel  3  is  at  Lx ;  that  the  maximum 
stresses  in  UJJ2  and  L2L3  are  each  135000  Ibs.  when  wheel  5  is 
at  Lo ;  and  that  the  maximum  stress  in  U2U3  is  145  ooo  Ibs.  when 
wheel  8  is  at  L3.  It  is  also  seen  that  wheel  7  at  L3  gives  almost 
as  large  a  stress  in  U2U3  as  does  wheel  8. 

If  a  truss  with  an  odd  number  of  panels  had  been  given,  then 
the  maximum  stress  in  the  center  panel  of  the  lower  chord  would 
have  occurred  when  the  loads  are  so  placed  that  there  would  be 
zero  shear  in  the  center  panel. 

If  the  live  load  had  consisted  of  two  engines  and  tenders  fol- 
lowed by  a  uniform  train  load,  then  the  stresses  caused  by  the 
loads  of  the  second  engine  should  also  have  been  considered. 

Web  Stresses.  The  diagram  for  all  the  web  stresses,  except 
the  hip  vertical,  is  shown  in  Fig.  148,  d,  and  for  the  hip  vertical, 
in  Fig.  148,  e.  For  the  diagram  shown  in  Fig.  148,  d,  the  pole 
is  taken  at  Px  with  a  pole  distance  equal  to  the  span  of  the  truss. 
When  wheel  3  is  placed  at  L±  (Fig.  148,  g),  the  left  reaction, 
which  is  represented  by  the  intercept  above  L0'  (Fig.  148,  g),— 
136000  Ibs. ;  the  shear  in  the  panel  L^  =136  ooo —  n  500  (the 
portion  of  the  loads  in  the  panel  L^  that  is  carried  to  L0  by  the 
stringer)  =  124  500  Ibs.;  and  the  stress  in  LoUj,  which  is  ob- 
tained by  drawing  a  line  parallel  to  the  member  'L0\Jlf=  153000 
Ibs.  (Fig.  148,  d).  This  is  the  maximum  stress  for  LoUj. ;  as 
may  be  shown  by  successively  placing  wheels  2  and  4  at  Llf  and 
finding  the  stresses  in  this  member  caused  by  these  positions  of 
the  loads.  Likewise,  the  maximum  stress  in  U^Lg  is  obtained 
when  wheel  3  is  at  L2  (Fig.  148,  h),  and  is  103  500  Ibs. ;  the  maxi- 
mum stresses  in  U,L2  and  U2L3  are  both  obtained  when  wheel  2 
is  at  Lo  (Fig.  148,  i),  and  are  50  700  and  62  300  Ibs.,  respectively; 
the  maximum  live  load  stresses  in  U3L3  and  U3L/  are  both 
obtained  when  wheel  2  is  placed  at  L/  (Fig.  148,  j),  and  are 
25  100  and  30900  Ibs.,  respectively  (if  there  is  no  dead  load  shear 
in  the  panel)  ;  and  the  maximum  live  load  stresses  in  U/L/ 


294  MAXIMUM    STRESSES — ENGINE    LOADS.        Chap.  XXIII. 

and  U/L/  are  both  obtained  when  wheel  2  is  at  L/  (Fig.  148,  k), 
and  are  6000  and  7300  Ibs.,  respectively  (if  there  is  no  dead  load 
shear  in  the  panel  L/L/).  When  wheel  2  is  at  any  panel  point, 
the  part  of  the  load  carried  to  the  panel  point  ahead  by  the 
stringer  is  4000  Ibs.  The  constructions  when  wheel  2  and  wheel  3 
are  successively  at  L3  are  shown  in  Fig.  148,  d,  wheel  2  at  L3 
giving  the  larger  stress. 

The  diagram  for  determining  the  maximum  stress  in  the  hip 
vertical  is  shown  in  Fig.  148,  e.  The  stress  in  this  member  is 
equal  to  that  portion  of  the  total  load  in  the  two  panels  L^  and 
LXL2  which  is  carried  to  Lt  by  the  stringers.  It  is  seen  that  this 
is  a  maximum  when  the  heavy  wheels  are  near  Lx.  To  determine 
the  exact  position  of  these  wheels  for  a  maximum  stress  in  UjL^ 
and  also  the  maximum  stress  in  the  member,  place  the  center  of 
gravity  of  the  first  six  wheels  (those  in  the  panels  L^  and  I^Lo) 
at  L!  ;  and  draw  the  funicular  polygon  shown  in  Fig.  148,  e,  with 
a  pole  distance  equal  to  one  panel  length.  If  the  pole  distance  is 
taken  equal  to  a  panel  length,  then  the  intercepts  in  the  funicular 
polygon  will  represent  the  reactions  at  L0  and  L2  caused  by  the 
loads  in  the  panels  L^  and  I^L.,,  respectively.  This  is  true 
because  in  the  funicular  polygon  intercept  X  pole  distance  (one 
panel  length)  =  moment;  and  the  reactions  =  intercept  X  pole 
distance  (one  panel  length)  -^moment  arm  (one  panel  length)  = 
intercept.  When  the  center  of  gravity  of  the  six  wheels  is  placed 
at  Lj,  the  reactions  at  L0  and  L2  are  each  equal  to  19  200  Ibs. 
(Fig.  148,  e),  and  the  stress  in  U^  =103  ooo  (the  total  weight 
of  the  six  wheels)  — 38400  =  64600  Ibs.  Now  place  wheel  3  at 
Lj.  For  this  position  of  the  loads,  the  reactions  at  L0  and  L2  are 
1 1  500  and  27  400  Ibs.,  respectively ;  and  the  stress  in  U^L!  = 
103000 —  ii  500  —  27400  =  64  100  Ibs.  Next  place  wheel  4  at 
L-L.  For  this  position  of  the  load,  an  additional  wheel — wheel  7 — 
is  brought  upon  the  length  L0L2 ;  the  reactions  at  L0  and  L2  are 
24  ooo  and  26  500  Ibs.,  respectively ;  and  the  stress  in  U^  = 
1 16  ooo  (the  total  weight  of  the  seven  wheels)  —  24  ooo  —  26  500 
~  65  500  Ibs.  It  is  thus  seen  that  the  maximum  tension  in  U^ 
=  65  500  Ibs.  when  wheel  4  is  at  L1B 

Referring  to  Fig.  148,  e,  it  is  seen  that  the  ordinates  under 


Art.  &.  LOAD  LINE  AND  MOMENT  DIAGRAM.  295 

wheels  2  and  3  represent  the  loads  that  were  deducted  from  the 
reactions  in  Fig.  148,  d  to  obtain  the  shears. 

The  stress  in  the  hip  vertical  may  be  obtained  in  a  different 
manner,  as  follows :  Place  wheel  4  at  L±,  and  draw  a  vertical 
through  wheel  4  to  intersect  the  funicular  polygon  at  v  (Fig. 
148,  b).  Also,  mark  the  points  m  and  n  on  the  funicular  polygon, 
distant  horizontally  one  panel  length  from  v.  The  lines  mv  and 
nv  are  the  closing  strings  of  the  funicular  polygons  for  the  loads 
in  the  panels  L0L±  and  I^Lo,  respectively.  Now  draw  rays 
through  P  parallel  to  these  closing  strings,  and  these  rays  will  cut 
off  on  the  load  line  the  total  load  transferred  to  L±,  which  is  the 
stress  in  U. 


ART.  2.     MAXIMUM  MOMENTS,  SHEARS,  AND  STRESSES  IN 
GIRDERS  AND  TRUSSES  OF  VARIOUS  TYPES  AND  SPANS. 

When  it  is  required  to  determine  the  moments  and  shears  in 
girders  and  trusses  of  different  spans,  subjected  to  the  same  load- 
ing, the  work  may  be  greatly  facilitated  by  constructing  a  load 
line  and  a  moment  diagram  for  the  load  on  one  rail.  If  the  dia- 
grams are  drawn  to  a  large  scale,  very  good  results  may  be 
obtained  by  their  use.  The  construction  of  the  diagrams  will  be 
explained,  and  then  their  application  to  the  determination  of 
moments  and  shears  in  girders  and  trusses  will  be  shown. 

217.  Load  Line  and  Moment  Diagram.  The  load  line  and 
moment  diagram  for  Cooper's  £40  loading  are  shown  in  Fig. 
149.  The  diagrams  shown  in  this  figure  were  originally  drawn 
upon  cross-section  paper,  divided  into  one-tenth  inch  squares, 
using  a  scale  of  four  times  that  shown  in  Fig.  149.  The  engine 
diagram  was  laid  off  to  a  scale  of  ten  feet  to  the  inch ;  the  loads, 
to  a  scale  of  50000  pounds  to  the  inch;  and  the  moments,  to  a 
scale  of  2  500  ooo  foot-pounds  to  the  inch.  The  engine  and  uni- 
form train  load  diagram  is  shown  in  the  lower  part  of  the  figure. 
The  engine  wheels  are  numbered,  the  load  on  each  wheel  shown, 
and  the  spacing  of  the  wrheels  given. 


296  MOMENTS   AND   SHEARS — ENGINE    LOADS.  Chap.  XXIII. 


S|S§S8S§§ 
FIG.  149.     MOMENT  AND  SHEAR  DIAGRAMS — COOPER'S  E40. 

The  load  line,  which  is  the  heavy  stepped  line  1-2-3-4-5,  etc., 
is  a  diagram  whose  ordinates  measured  from  the  line  o-o  repre- 


Art.2,  LOAD   LINE   AND   MOMENT   DIAGRAM.  297 

sent  the  summation  of  the  loads  to  the  left.  Each  step  in  the 
load  line  represents  to  scale  the  load  directly  under  it.  The  por- 
tion of  the  diagram  above  the  uniform  load  is  a  straight  line 
having  a  uniform  slope  of  2000  pounds  per  foot. 

The  moment  lines,  which  are  numbered  i,  2,  3,  4,  5,  etc.,  near 
the  right  edge  of  the  figure,  are  constructed  as  follows :  Starting 
at  o,  the  right  end  of  the  horizontal  reference  line  o-o,  lay  off 
successively  on  a  vertical  line  the  moment  of  each  wheel  load 
about  that  point,  beginning  with  wheel  i.  The  moment  line  i  is 
then  drawn  from  a  point  in  the  reference  line,  directly  over 
wheel  i,  to  the  first  point  at  the  right  of  the  diagram;  moment 
line  2  is  drawn  from  a  point  on  moment  line  I,  directly  over 
wheel  2,  to  the  second  point  at  the  right  of  the  diagram;  moment 
line  3,  from  a  point  on  moment  line  2,  directly  over  wheel  3,  to 
the  third  point,  etc.  The  moment  line  BC  is  a  parabolic  curve; 
and  may  be  easily  constructed  by  computing  the  moments  of  por- 
tions of  the  uniform  load  between  B  and  several  points  to  the 
right  about  these  points,  and  laying  off  these  moments  above  the 
moment  line  18.  It  is  seen  that  the  broken  and  curved  line  ABC 
is  an  equilibrium  polygon  for  the  given  loads.  The  ordinate  at 
any  point,  measured  between  the  reference  line  o-o  and  any 
moment  line,  represents  the  sum  of  the  moments  about  this  point 
of  all  the  loads  up  to  and  including  the  wheel  load  corresponding 
to  the  moment  line  under  consideration.  To  illustrate,  the  sum 
of  the  moments  of  the  wheel  loads  I,  2,  3,  and  4  about  wheel  18 
is  represented  by  the  ordinate  over  wheel  18,  measured  between 
the  reference  line  o-o  and  moment  line  4.  Also,  the  moment  at 
any  point,  measured  between  any  two  moment  lines — as  between 
2  and  6 — represents  the  sum  of  the  moments  of  wheel,  loads  3  to 
6,  inclusive,  about  that  point.  The  line  of  equal  shears  for  wheels 

P, 
i  and  2,  shown  in  Fig.  149,  has  an  upward  slope  of-r-;   where 

P!  is  the  first  wheel  load,  and  b  is  the  distance  between  wheels 
i  and  2.  The  use  of  this  line  will  be  explained  in  §  220  ( i ) . 

218.  Application  of  Diagrams  in  Fig.  149  to  Determining 
Maximum  Moments  in  Plate  Girders  or  at  Joints  of  the 


298 


MAX.  MOMENTS  AND  SHEARS — ENGINE  LOADS.      Chap.  XXIII. 


Loaded  Chord  of  a  Truss  with  Parallel  or  Inclined  Chords. 
The  diagrams  shown  in  Fig.  149  may  be  used  for  finding  the 
position  of  the  wheels  for  a  maximum  moment,  and  also  the  value 
of  the  moment.  To  illustrate  the  use  of  these  diagrams,  let  it  be 
required  to  determine  the  position  of  the  wheels  for  a  maximum 
moment  at  L2  (Fig,  150),  this  point  being  either  a  panel  point  of 


u,  u*  iS-J7-''        l 

A  ~t^T7 


Maximum  Moment  at 


Lz 


FIG.  150.     MAXIMUM  MOMENT  AT  LOADED  CHORD  JOINT. 

the  truss  whose  span  is  AB,  or  any  point  along  a  girder  of  the 
same  span.  To  avoid  confusion,  a  portion  of  the  load  line  shown 
in  Fig.  149  is  reproduced  to  a  larger  scale  in  Fig.  150.  Since  the 
girder,  or  truss,  must  be  shifted,  the  points  along  the  girder,  or 
the  panel  points  of  the  truss,  should  be  marked  on  the  edge  of 
a  separate  slip  of  paper  or  upon  a  piece  of  tracing  cloth ;  and  it 
will  be  assumed  that  the  points  marked  on  the  span  AB  (Fig. 
150)  are  on  a  separate  slip  of  paper.  The  criterion  for  a  maxi- 
mum moment  at  L2  is  —  the  average  load  2  Pt  to  the  left  of  L2 
must  be  equal  to  the  average  load  2  P  on  the  entire  span  (see 
§  208).  Try  wheel  4  at  L2,  i.  e.,  shift  the  truss  until  L2  is  under 
this  wheel  (Fig.  150).  The  total  load  on  the  span  is  represented 

2  P 
by  the  ordinate  BC  =  2  P;  while  the  average  load  is  — ^ — ,  which 

is  represented  by  the  slope  of  the  line  AC.  The  load  S  P!  to  the 
left  of  L,  is  represented  either  by  the  ordinate  FE  or  the  ordinate 
FD,  depending  upon  whether  wheel  4  is  at  L2  or  just  to  the  left 


Art.  2.  MAX.  MOMENT  AT  LOADED  CHORD  JOINT.  299 

SPj 

of  L2 ;   and  the  average  load    "iTF"*0  tne  kft  °f  L2  is  represented 

either  by  the  slope  of  the  line  AE  or  the  line  AD.  Since  the  slope 
of  the  line  AC  is  less  than  that  of  AD  and  greater  than  that  of 
AE,  it  is  seen  that  wheel  4  at  L2  gives  a  maximum  moment  at 
this  point.  If  the  line  AC  lays  above  the  line  AD,  the  loads  must 
be  moved  to  the  left;  and  if  the  line  AC  is  below  the  line  AE, 
then  the  loads  must  be  moved  to  the  right.  It  is  thus  seen  that 

2P 
if  the  line  whose  slope  is  — - —  cuts  the  vertical  line  representing 

J—/ 

the  load  at  the  point,  this  position  of  the  load  gives  a  maximum 
moment.  Instead  of  actually  drawing  the  line  AC,  its  position  is 
usually  determined  by  stretching  a  thread.  In  the  illustration  just 
given,  none  of  the  wheels  were  off  the  bridge  to  the  left,  and  the 
line  AC  starts  from  the  zero  shear  line  at  L0.  If  some  of  the 
loads  had  passed  the  left  end  of  the  bridge  L0,  then  AC  should 
start  in  the  load  line  vertically  over  L0. 

The  position  of  the  wheels  having  been  determined,  the 
moment  itself  may  be  easily  found  from  the  moment  diagram 
(Fig.  149),  as  follows:  With  wheel  4  at  L2,  read  the  ordinate 
at  the  right  end  of  the  span  L0'  between  the  reference  line  o-o 
and  the  moment  line  10,  which  is  the  moment  of  all  the  loads  on 
the  span  about  L0'.  The  moment  at  L2  is  obtained  from  the 

AF 

.moment  at  L0'  by   multiplying  it  by    T —  and   subtracting  the 

JL/ 

moment  of  the  loads  to  the  left  of  L2,  this  latter  moment  being 
given  by  reading  the  ordinate  at  L2  between  the  reference  line 
o-o  and  moment  line  4. 

Since  the  line  ABC  (Fig.  149)  formed  by  the  segments  of 
the  moment  lines  is  a  funicular  polygon  for  the  given  loads,  the 
moment  at  L2  is  also  equal  to  the  ordinate  at  that  point  inter- 
cepted between  the  funicular  polygon  and  the  closing  line.  The 
extremities  of  this  closing  line  are  on  the  verticals  which  pass 
through  the  ends  of  the  bridge. 

In  finding  the  maximum  moment  in  a  girder — which  occurs 
near  its  center — it  is  necessary  to  locate  the  center  of  gravity  of 


300 


MAX.  MOMENTS  AND  SHEARS — ENGINE  LOADS.      Chap.  XXIII. 


all  the  loads  on  the  girder  (see  §  159  (a))  ;  and  the  center  of 
gravity  of  any  number  of  loads  may  be  found  from  Fig.  149  by 
producing  the  extreme  strings  of  the  funicular  polygon  until  they 
intersect. 

219.  Application  of  Diagrams  in  Fig.  149  to  Determining 
Maximum  Moments  at  Panel  Points  in  the  Unloaded  Chord 
of  a  Truss  with  Parallel  or  Inclined  Chords.  The  use  of  the 
diagrams  in  Fig.  149  for  finding  the  maximum  moment  at  any 
unloaded  chord  panel  point  will  be  shown  by  the  following  prob- 
lem :  It  is  required  to  determine  the  position  of  the  engine  load 
for  a  maximum  moment  at  the  panel  point  U3  of  the  truss  shown 


Maximum  Moment  at  Us 


PIG.  151.     MAXIMUM  MOMENT  AT  UNLOADED  CHORD  JOINT. 

in  Fig.  151,  together  with  value  of  the  maximum  moment.  Let 
the  heavy  stepped  line  1-2-3-4,  etc.  (Fig.  151),  be  a  portion  of 
the  load  line,  and  let  L  be  the  span  of  the  truss.  It  has  been 
shown  (§  209)  that  the  criterion  for  a  maximum  moment  at  U3  is 


+  2  Pt 


;  where  2  P  is  the  total  load  on  the  span, 


2  Pj  is  the  load  to  the  left  of  L2,  2  P2  is  the  load  in  the  panel 
L2L/5  d  is  the  panel  length,  r  is  the  horizontal  distance  from  U3 
to  L2,  and  s  is  the  horizontal  distance  from  U:,  to  L0.  Try 
wheel  6  at  L,',  which  is  the  position  of  the  load  in  Fig.  151.  Now 
the  total  load  S  P  on  the  span  is  represented  by  BC,  and  the  aver- 


Art.  2.  MAX.    MOMENT    AT    UNLOADED    CHORD    JOINT.  301 

2P 

age  load  —  =-•  ,  by  the  slope  of  the  line  AC.    The  load  2  Pt  to  the 

left  of  Lo  is  represented  by  GK,  and  that  in  the  panel  L2L./,  by 
HD  or  HE  (depending  upon  whether  wheel  6  is  just  to  the  left 
or  to  the  right  of  L2')-  The  average  load  in  the  panel  L2L2'  is 
represented  by  the  slope  of  the  line  GD,  or  the  line  GE.  The 

r 
term   2  P2—  r(see   above    criterion)    is    represented    by   JN,   or 

JM  ;    the  term  2  P2  -+  2  Plf  by  RN,  or   RM  ;    and   the  term 


,  by  the  slope  of  the  line  AN,  or  the  line  AM   (not 


drawn  in  the  figure).  Since  the  line  AC  cuts  the  vertical  through 
U3  between  the  points  N  and  M,  it  is  seen  that  the  wheel  6  at  L/ 
satisfies  the  criterion  for  a  maximum  moment  at  U3.  If  it  had 
been  impossible  to  satisfy  the  criterion  by  placing  some  wheel  at 
Lo',  then  the  criterion  should  have  been  tested  by  placing  a  wheel 
at  Lo.  In  using  the  diagrams  shown  in  Fig.  149,  the  upper  and 
lower  chord  panel  points  should  be  marked  on  a  separate  slip  of 
paper.  The  slip  should  then  be  shifted  until  the  wheel  which 
gives  a  maximum  moment  is  at  the  panel  point  about  which  the 
moment  is  required.  The  auxiliary  lines  shown  in  Fig.  151  need 
not  be  drawn  ;  as  the  positions  of  the  lines  GD  and  GE  may  be 
determined  by  stretching  a  thread,  the  points  N  and  M  being 
marked,  and  the  thread  then  moved  to  the  position  AC. 

The  position  of  the  engine  load  (wheel  6  at  L2')  having  been 
determined,  the  left  reaction  may  be  found  as  in  §  218.  The 
determination  of  the  moment  at  the  unloaded  chord  joint  U3 
differs  from  that  at  a  loaded  chord  joint,  in  that  only  the  portion 
of  the  load  in  the  panel  L2L2',  transferred  to  L/  by  the  stringers, 
should  be  considered.  The  moment  at  U3  is  equal  to  that  of  the 
left  reaction  minus  the  moment  of  the  portion  of  the  loads  in 
LXo'  transferred  to  L2  together  with  the  moment  of  all  other 
loads  to  the  left  of  U3. 


302          MAX.  MOMENTS  AND  SHEARS  -  ENGINE  LOADS.        Chap.  XXIII. 

The  moment  at  U3  may  also  be  readily  found,  as  follows: 
With  wheel  6  at  L/,  find  the  moments  at  L2  and  L/,  as  in  §  218. 
Then  if  MLand  MB  represent  the  moments  at  L2  and  L/  respect- 


ively,  the  moment  at  U3  = 


(MK  —  ML)  -7- 


220.  Application  of  Diagrams  in  Fig.  149  to  Determining 
Maximum  Shears.  Two  cases  will  be  considered,  viz  : 
(i)  maximum  shears  in  beams  and  girders;  and  (2)  maximum 
shears  in  trusses. 

(i)  Maximum  Shears  in  Beams  and  Girders.  It  has  been 
shown  (§  21  1  )  that  wheel  i  at  any  point  will  give  a  maximum 

P       3  P 
shear  when  —  r-==^f  —  >  and  that  wheel  2  will  give  a  maximum 

D     >     L*t 

PI      2P± 
shear  when  —  "  —  =-  ;   where  Px  is  the  first  wheel  load,  S  P!  is 

D          ^  JLr 

the  total  load  on  the  span  when  Px  is  at  the  point,  2,  P2  is  the  total 
load  when  P2  is  at  the  point,  b  is  the  distance  between  wheels  i 
and  2,  and  L  is  the  span  of  the  beam  or  girder. 


5 

4 

-^.*-*"<-' 

I3 

^•^'^ 

b     J  -—  *" 

Maximum  Shears  in  Girder 

M  |L-^-      ;" 

N 

* e — £ u B 

FIG.  152.     MAXIMUM  SHEARS  IN  A  GIRDER. 

Let  1-2-3-4,  etc.  (Fig.  152)  be  a  portion  of  the  load  line, 
and  let  AB  =  L  be  the  length  of  a  beam  or  girder.  It  is  required 
to  determine  the  segment  of  the  beam  in  which  wheel  i  gives  the 
maximum  shear,  and  also  the  segment  in  which  wheel  2  gives  the 
maximum  shear.  Place  wheel  i  at  the  left  end  of  the  beam ;  and 


<4rt.£.  MAXIMUM    SHEARS    IN    A   GIRDER.  303 

p 

from  A,  draw  the  line  AC  having  a  slope  of -7—  to  intersect  a  ver- 
tical through  the  right  end  of  the  beam.  Now  the  ordinate  BC 

p 
represents  the  load  which  divided  by  L  equals -r^-.     It  therefore 

represents  the  load  S  Plt  or  2  P2.  Draw  the  line  CD  parallel  to 
AB  to  intersect  the  load  line  at  D;  and  also  draw  the  vertical 
line  DE.  It  is  then  seen  that  the  average  load  on  the  entire  span 

Pt 
equals—,  when  wheel  5  is  at  the  right  end  of  the  beam.    Now  lay 

off  the  span  AB  on  a  separate  slip  of  paper,  and  mark  the  points 
E'  and  F'  on  this  slip  to  correspond  with  the  points  E  and  F 
(Fig.  152).  Place  the  point  B  (on  the  slip)  at  E,  i.e.,  directly 
under  wheel  5,  and  mark  the  positions  of  wheels  I  and  2,  calling 
these  points  i'  and  2'.  Then  any  point  in  the  segment  2'B 
(between  wheel  2  and  the  right  end  of  the  span)  has  its  maxi- 
mum shear  when  wheel  i  is  at  the  point ;  and  any  point  to  the 
left  of  i'  (wheel  i)  has  its  maximum  shear  when  wheel  2  is  at 
the  point.  Between  i'  and  2'  (the  distance  between  wheels  I 
and  2),  both  positions  should  be  tried. 

The  line  marked  equal  shears,  wheels  i  and  2  (Fig.  149), 
corresponds  to  the  line  AC  (Fig.  152).  Since  Fig.  149  has  cross- 
section  lines,  it  is  unnecessary  to  draw  any  additional  lines. 

(2)  Maximum  Shears  in  Trusses.  It  has  been  shown  (§212) 
that  the  criterion  for  a  maximum  shear  in  any  panel  of  a  truss 
is — the  load  in  the  panel  must  be  equal  to  the  total  load  on  the 
span  divided  by  the  number  of  panels.  If  2  P  is  the  total  load 
on  the  span,  and  3  P±  is  the  load  in  the  panel  other  than  the 
wheel  load  at  the  panel  point  to  the  right,  then  any  load  P  at 

2P 
this  panel  point  will  give  a  maximum  shear  in  the  panel  it-= — 

2  P,  2  P!  +  P 

lavs  between—  : —  and  : . 

d  d 

Let  L  (Fig.  153)  be  the  span  of  any  truss,  and  1-2-3-4,  etc., 
a  portion  of  the  load  line.  It  is  required  to  determine  which 


304 


MAX.  MOMENTS  AND  SHEARS — ENGINE  LOADS.      Chap.  XXIII. 


wheel  load  placed  at  the  panel  point  to  the  right  will  give  a  maxi- 
mum shear  in  each  panel  of  the  truss.  Place  wheel  i  at  the  first 
panel  point  L1?  as  shown  in  Fig.  153.  Draw  the  lines  AC,  AD, 


U, _U2_  _Uk 

/IT  ~7*7\  ~ 
/     X     !--.H"'' 

/  krX^T&E5- 

'  "-  :™-^  FT/ \ 


.L.  _U_H      L_          Liz  _L, 

FIG.  153.     MAXIMUM  SHEARS  IN  A  TRUSS. 

and  AK  with  ordinates  at  L±  representing  the  wheel  loads  P±,  P2, 

P  '•  P  +  P 

and  P3,  respectively.    These  lines  will  have  slopes  of  — :-> — - — 

d        d 


and 


•,  respectively.     Wheel  i  placed  at  any  panel 


point  will  give  a  maximum  shear  in  the  panel  to  the  left  when 

Pt        2P  SP 

— T-  =  -T —  Now-^ — is  represented  by  BC,  the  ordinate  over  the 
d     >    L.  L, 

right  end  of  the  truss  L/;    and  -:-==  — —  as  soon  as  wheel  3 

enters  the  span  and  until  wheel  4  enters  the  span  at  L0'.  There- 
fore wheel  i  will  give  a  maximum  shear  in  the  panels  to  the  left 
of  each  panel  point  passed  by  it  in  moving  the  loads  to  the  left 
until  wheel  4  enters  the  span  at  L/,  i.  e.,  in  the  panel  L/L/. 
Likewise,  wheel  2  will  give  a  maximum  shear  in  any  panel  when 

-=—  lays  between-^- and  — -= ;   i.  e.,  for  values  of  2  P  between 

d  d 

BC  and  BD.     Now 


Therefore  wheel  2  will  give  a  maximum  shear  in  the  panels  to 


Art. 2.  MAXIMUM    SHEARS   IX    TRUSSES.  305 

the  left  of  each  panel  point  passed  by  it  in  moving  the  loads  from 
the  position  with  wheel  4  at  L0'  to  the  position  with  wheel  10  at 
L0',  i.  e.,  in  the  panels  L/L/,  L2L/,  and  L^L,,.  Likewise,  wheel  3 
will  give  a  maximum  shear  in  the  panels  to  the  left  of  each  panel 
point  passed  by  it  in  moving  the  loads  from  the  position  with 
wheel  10  at  L0'  to  the  left  end  of  the  truss,  i.  e.,  in  the  panels 
L0Lj_  and  I^Lo.  This  is  determined  from  the  slopes  of  the  lines 
AD  and  AK.  From  the  above  discussion,  it  is  seen  that  both 
wheel  i  and  wheel  2.  satisfy  the  criterion  for  a  maximum  shear 
in  the  panel  L/L/ ;  and  that  both  wheel  2  and  wheel  3  sat- 
isfy the  criterion  for  a  maximum  shear  in  the  panel  LxLo. 
Since  the  diagram  in  Fig.  149  has  cross-section  lines,  the 
auxiliary  lines  shown  in  Fig.  153  need  not  be  drawn,  the 
actual  process  being  as  follows :  Lay  off  the  span  AB  =  L 
on  the  edge  of  a  separate  slip  of  paper,  and  mark  the  panel 
points  L0,  L!,  L2,  L/,  L/,  and  L0'.  Place  the  edge  of  the 
slip  on  the  reference  line  o-o  (Fig.  149)  with  the  right  end  L0 
of  the  truss  under  wheel  4,  and  mark  the  positions  of  wheel  i 
and  wheel  2.  on  the  slip.  Now  move  the  slip  to  the  right  until 
L0'  is  directly  under  wheel  10,  and  mark  the  positions  of  wheel  2. 
and  wheel  3.  The  marking  of  the  slip  will  now  correspond  to 
that  of  the  line  AB  (Fig.  153).  It  is  now  seen  that  wheel  I  will 
satisfy  the  criterion  for  a  maximum  shear  in  the  segment  L0'i'; 
that  wheel  2  will  satisfy  the  criterion  for  a  maximum  shear  in 
the  segment  2f2f ;  and  that  wheel  3  will  satisfy  the  criterion,  in 
the  segment  3'L0.  It  is  also  seen  that  the  segment  in  which 
wheel  i  satisfies  the  criterion  for  a  maximum  shear  is  overlapped 
by  that  in  which  wheel  2  satisfies  the  criterion,  by  the  distance 
i  '2',  which  is  the  distance  between  wheels  i  and  2.  It  is  further 
seen  that  the  segment  in  which  wheel  2  satisfies  the  criterion  for 
a  maximum  shear  is  overlapped  by  that  in  which  wheel  3  satisfies 
the  criterion,  by  the  distance  between  wheels  2  and  3. 

The  position  of  the  wheels  having  been  determined,  the  shear 
itself  may  be  easily  found  from  the  moment  lines  (Fig.  149). 
To  determine  the  shear  in  any  panel,  say  in  LX/  (Fig.  153), 
place  the  right  panel  point  L/  under  wheel  2  (this  position  of 


306 


MAXIMUM   STRESSES ENGINE  LOADS. 


Chap.  XXIII. 


the  load  being  the  one  which  satisfies  the  criterion  for  a  maximum 
shear)  ;  and  read  the  ordinate  to  the  funicular  polygon  ABC 
at  the  right  end  of  the  truss.  The  moment  represented  by  this 
ordinate  divided  by  the  span  is  equal  to  the  left  reaction.  Since 
the  loads  can  only  come  upon  the  truss  at  the  panel  points,  being 
transferred  by  the  stringers,  the  shear  in  the  panel  L2L/  is  equal 
to  the  left  reaction  minus  the  portion  of  the  load  in  this  panel 
which  is  carried  to  the  left  panel  point  L2.  The  load  which  is 
carried  to  this  panel  point  is  equal  to  the  moment  of  the  load  to 
the  left  of  L/  about  L,'  divided  by  the  panel  length  d ;  and  is 
represented  by  the  ordinate  to  the  funicular  polygon  above  the 
panel  point  L/. 

221.  Application  of  Diagrams  in  Fig.  149  to  Determining 
Maximum  Web  Stresses  in  Trusses  with  Inclined  Chords. 
The  following  problem  will  show  the  application  of  the  diagrams 
in  Fig.  149  to  determining  the  maximum  stress  in  any  web  mem- 


Uz 


FIG.  154.     MAXIMUM  STUESS  IN  WEB  MEMBER — TRUSS  WITH  INCLINED  CHORDS. 


ber  of  a  truss  with  inclined  chords.  It  is  required  to  find  the 
maximum  live  load  stress  in  the  member  U^L2  of  the  truss  shown 
in  Fig.  154.  A  portion  of  the  load  line  shown  in  Fig.  149  is 
reproduced  to  a  larger  scale  in  Fig.  154.  It  has  been  shown 
(§  213)  that  the  criterion  for  a  maximum  stress  in  the  member 


U.L,  is  2_  = 
JU 


••Hi). 


;   where  2  P  is  the  total  load  on  the 


Art.S.  MAX.  WEB  STRESSES INCLINED  CHORDS.  307 

span,  2  r\  is  the  load  in  the  panel  I^Lo,  s  is  the  distance  from  Lx 
to  the  left  end  of  the  truss,  and  k  is  the  distance  from  the  left  end 
of  the  truss  to  the  point  of  intersection  O  of  the  members  UJJo 
and  LjLo,  this  point  being  the  center  of  moments  for  determining 
the  stress  in  L^L,,.  It  is  seen  that  this  criterion  is  similar  to  that 
for  a  maximum  shear  in  the  panel,  indicating  that  some  wheel 
near  the  head  of  the  train  placed  at  L2  will  give  a  maximum 
stress  in  l^L,,.  Try  wheel  2  at  L2  (Fig.  154).  The  load  2  Pt 
in  the  panel  LXL2  is  either  Pt  or  P±  +  P2,  depending  upon 
whether  wheel  2  is  to  the  right  or  to  the  left  of  L2.  The  term 

3  p  ( i  -|-  ~  J  of  the  criterion  is  represented  by  GF,  or  by  GE; 


and  the  term : ,  by  the  slope  of  the  line  AF,  or  the 

line  AE.  The  total  load  on  the  span  is  represented  by  BC,  and 
the  average  load,  by  the  slope  of  the  line  AC.  Since  AC  lays 
between  AF  and  AE,  it  is  seen  that  wheel  2  satisfies  the  criterion 
for  a  maximum  stress  in  UjL,. 

To  determine  the  stress  in  the  member  UjLo  place  wheel  2 
at  L2,  and  read  the  ordinate  to  the  funicular  polygon  (Fig.  149) 
at  the  right  end  of  the  truss.  The  moment  represented  by  this 
ordinate  divided  by  the  span  L  is  equal  to  the  left  reaction  R18 
If  G!  represents  the  portion  of  the  load  to  the  left  of  the  section 
p-p  (i.  e.,  the  portion  of  the  load  carried  to  Lx  by  the  stringer), 

Rjk  — d  (k  +  s) 

then  the  stress  in  U1L2  = =- .     The  part  of  the 

load  in  LXL2  which  is  carried  to  Lx  may  be  found  by  reading  the 
ordinate  to  the  funicular  polygon  at  L2,  and  dividing  by  the  panel 
length. 

222.  Determination  of  Maximum  Stresses  in  a  Truss  with 
Subordinate  Bracing.  Petit  Truss.  A  truss  with  subordinate 
bracing  has  been  defined  as  one  which  has  points  of  support  for 
the  floor  system  between  the  main  panel  points.  The  Baltimore 
trusses  shown  in  Fig.  114,  d  and  Fig.  114,  e  are  examples  of  such 


308 


MAXIMUM    STHESSES — ENGINE    LOADS.  Chap.  XXIII. 


a  truss,  in  which  the  chords  are  parallel;  while  the  Petit  trusses 
shown  in  Fig.  114,  j  and  Fig.  155  are  examples  of  this  type,  in 
which  the  chords  are  not  parallel.  The  effect  of  the  subordinate 
bracing  upon  the  stresses  in  the  main  members  of  the  truss  will 
be  shown  by  the  following  problem:  It  is  required  to  determine 
the  maximum  stresses  in  the  members  in  the  panel  L4L6  of  the 
truss  shown  in  Fig.  155.  The  members  L6M,  U2M,  and  L5M 
are  tension  members;  while  L4M  is  in  compression  when  the 
counter  U3M  is  not  acting,  and  in  tension  when  the  counter  is 
acting.  It  is  seen  that  U3M  does  not  act  when  the  main  members 
have  their  maximum  stresses. 


U 


FIG.  155.     MAXIMUM  STRESSES — TRUSS  WITH  SUBORDINATE  BRACING. 

To  determine  the  maximum  stress  in  U2U3,  cut  the  members 
U2U3,  L6M,  and  L5L6  by  the  section  p-p,  and  take  the  center  of 
moments  at  L6.  Since  the  counter  U8M  is  not  acting,  the  posi- 
tion of  the  loads  for  a  maximum  moment  at  L6,  together  with  the 
value  of  this  moment,  may  be  obtained  by  the  methods  shown 
in  §  218.  The  stress  in  U2U3  is  equal  to  this  moment  divided  by 
the  perpendicular  distance  from  L6  to  U2U3.  It  is  thus  seen  that 
the  stress  in  this  chord  member  is  the  same  as  for  a  truss  with 
the  subordinate  bracing  omitted. 

To  find  the  maximum  stress  in  the  lower  chord  member  L4L6, 
the  same  section  should  be  taken  with  the  center  of  moment  at  U.,. 
The  position  of  the  wheel  loads  for  a  maximum  moment  at  U2 
may  be  determined  from  the  criterion  deduced  in  §  210.  By  the 
use  of  the  diagrams  in  Fig.  149,  it  is  easy  to  apply  this  criterion, 
and  to  determine  which  wheel  placed  at  L5  will  give  a  maximum 
moment  at  U2.  The  moment  at  this  panel  point  is  equal  to  the 


Art.  2.          TRUSS  WITH  SUBORDINATE  BRACING.  309 

reaction  at  L0  multiplied  by  the  distance  L0L4  minus  the  moment 
of  the  loads  to  the  left  of  L4  about  L4  plus  the  moment  of  the 
portion  of  the  loads  in  the  panels  L4L5  and  L5L6  which  is  carried 
to  L5  by  the  stringers.  It  is  necessary  to  consider  the  joint  load 
at  L5 ;  since  it  is  to  the  left  of  the  section  p-p. 

The  maximum  stress  in  L5M  is  obtained  when  there  is  a  maxi- 
mum joint  load  at  L5,  and  is  equal  to  that  load.  The  stress  in 
this  member  may  be  obtained  as  shown  in  Fig.  148,  e,  and  ex- 
plained in  §  216. 

The  position  of  the  wheel  loads  for  a  maximum  stress  in  U2M 
and  the  stress  itself  are  the  same  as  if  the  subordinate  members 
L5M  and  L4M  were  omitted.  The  stress  in  U2M  may  therefore 
be  determined  as  shown  in  §  221,  the  subordinate  members  being 
omitted,  and  the  panel  length  taken  as  L4L6.  It  is  seen  that  this 
is  true,  since  the  small  triangular  truss  L4ML6  merely  acts  as  a 
trussed  stringer  to  transfer  the  loads  to  L4  and  L6. 

The  stress  in  L0M  is  influenced  by  the  subordinate  members 
if  there  are  any  loads  between  L4  and  L6.  The  stress  in  this 
member  may  be  determined  by  taking  the  center  of  moments  at 
the  point  of  intersection  of  U2U3  and  L4L6. 

The  maximum  stress  in  U2L4  may  be  determined  as  in  §  221, 
considering  the  members  L6M  and  L4M  removed.  The  section 
r-r  should  be  cut,  and  the  center  of  moments  taken  at  the  point 
of  intersection  of  UJJ2  and  L4L6. 

The  stress  in  L4M  when  the  counter  is  not  acting,  i.  e.,  when 
L4M  acts  as  a  compression  member,  may  be  readily  found  by 
graphic  resolution.  Since  U2M  and  L6M  are  collinear,  the  re- 
solved components  in  L4M  and  L5M  perpendicular  to  U2L6  must 
be  equal. 

The  maximum  tensile  stress  in  L4M,  which  occurs  when  the 
counter  U3M  is  acting,  may  be  found  in  the  following  manner : 
Consider  the  numbers  L4M  and  U3M  replaced  by  a  straight  mem- 
ber L4U3.  Now  the  maximum  stress  in  this  member  may  be 
found  by  the  same  method  used  for  that  in  U2M,  i.  e.,  by  consid- 
ering the  members  L5M  and  U2M  removed.  The  mehiber  L6M 
is  not  considered ;  as  it  does  not  act  for  this  loading.  After  find- 
ing the  stress  in  L4U3,  the  maximum  stress  in  L4M  may  be  deter- 


310  MAXIMUM    STRESSES — ENGINE    LOADS.        Chap.  XXIII. 

mined,  as  follows:  Lay  off  on  a  line  parallel  to  the  member 
L4U3  a  length  L4U3,  representing  to  scale  the  stress  in  that  mem- 
ber. Through  U3,  draw  a  line  parallel  to  the  member  L6M ;  and 
through  L4,  draw  lines  parallel  respectively  to  L4M  and  U3M  to 
intersect  the  line  parallel  to  L6M.  This  construction  gives  the 
maximum  stress  in  the  member  L4M. 

The  maximum  stress  in  the  counter  U3M  may  be  found  by 
taking  the  section  p-p,  remembering  that  the  member  L6M  is  not 
acting  when  the  truss  is  loaded  for  a  maximum  stress  in  the 
counter.  The  load  should  be  brought  upon  the  truss  at  L0,  and 
the  left  segment  loaded  for  a  maximum  in  the  member.  The 
hanger  L5M  should  be  considered;  as  the  loads  carried  by  it 
increase  the  stress  in  UaM. 


INDEX. 


PAGE. 

Accurate  method,    Moment   of   In- 
ertia      61 

Action,  known  lines  of 23 

Line   of,    defined 5 

Algebraic  formulae  for  beams 198 

methods    for    beams 167 

moments,  stresses  by 86 

resolution,   stresses   by 95 

Application  of  diagrams 297 

web   stress 306 

Point  of,  defined 5 

Approximate    method,    moment    of 

inertia 59 

Arch,  defined 143 

Line  of  pressure  in 35,  37 

Area,  center  of  gravity  of,  47  ;  ir- 
regular      50 

Irregular   49 

Geometrical    47 

Moment     39 

Moment  of  inertia  of 58 

about   parallel   axes 64 

of  parallelogram 47 

of    quadrilateral 48 

of    sector 48 

of    segment 48 

of    triangle 48 

Radius  of  gyration  of GO 

Axes,   parallel,   moment   of  inertia 

of   .,  56 


Baltimore  bridge  truss 210 

Coefficients    for 264 

Base  of  column 151 

wind  load  stress,  hinged  column  155 

Beam,    cantilever,    moment,    deflec- 
tion and  shear 188 

fixed   both   ends 195 

Floor,  maximum  reaction 279 

Overhanging,  concentrated  loads  175 

Overhanging,   uniform   load 177 

one  end  fixed 189 

Beams 167 

Deflection   in 178 

Restrained    187 

Bending  moment,  see  also  Moment. 

in    beams 167 

in  simple  beam 171 

Bent,  stresses  in  transverse 150 

Trestle    162 

Body,  rigid,  defined 3 


PAGE. 

Bow's  notation 86 

Bowstring  bridge  truss 210 

truss,  stresses  in 243,  247 

Braces,  main 127 

Bracing,  bridge 212 

Subordinate 273,  308 

Sway 70 

Weight    of 70 

Bridge,  see  also  Trusses. 

Bridge,  floor  system 213 

joists    213 

loads    214 

stringers 213 

truss    members 211 

trusses,   see   names   of   trusses, 

trusses,  types  of 210 

Bridges   209 

Live  loads  for 216 

Wind  load  on , 219 

Weights  of 214 

Building,  transverse  bent 150 


Camel's  back  bridge  truss 210 

Cantilever   beam 168 

beam,  moment,  deflection,  shear  188 

roof    truss 67 

trusses    114 

Center  of  gravity 47 

irregular    area 50 

Centroid,  47;  of  parallel  forces...  49 

Chord,    defined 68 

Loads  on  upper,  99;  on  lower..  103 

Maximum  stress  in,  Pratt  truss.  291 
Chords,      counter-braced,      parallel, 

139  ;    non-parallel 141 

Inclined,  306;  truss  with 280 

Loaded    298 

Stresses  in  trusses  with  parallel  259 

Circular  chord  truss 67 

Closed    polygon 11 

Closing  funicular  polygon 23 

Coefficients,  method  of  for  stresses  259 

for  Baltimore  truss 264 

for  Pratt  truss 264 

for  Warren  truss 264 

Columns,  conditions  of  ends 150 

fixed  at  base 151,  152,  157 

fixed  at  top 152 

hinged  top  and  base 151 

hinged  at  base,  stresses  in 155 

Combined  stress  diagram 116 

counterbraced    truss 139 

Complete  frame  structure 65 


311 


312 


INDEX. 


PAGE. 

Components,    defined,    6 ;    horizon- 
tal      80 

horizontal,   or  reactions  equal..  108 
Non-parallel,   non-concurrent. 23,  24 
Composition  of  forces,   6  ;  concur- 
rent  forces 8 

Compression,    defined 84 

members,    long 68 

Compressive  stress,  sign  of 86 

Concentrated       loads,       cantilever 

beam    168 

one  end  fixed 189 

overhanging    beam 175 

simple    beam. 171 

moving  load 201 

wheel   loads 217 

Concurrent   forces 5,  8 

Equilibrium    of 12 

Resolution  of 11 

Resultant    of 9 

Conditions  for  equilibrium 26 

of  ends  of  columns 150 

Connections    213 

Eccentric    riveted 164 

Constant  moment  of  inertia 178 

Construction,  special,  for  funicular 

polygon 34 

of  a   roof 69 

Simple,  beam  one  end  fixed....  191 

Contraction  and  expansion 70 

in     bridges 213 

Cooper's  Class  E-40 296 

Copianar  forces  defined 5 

Corrugated  steel,  69;  weight  of..  70 
Counterbraced       truss,       combined 

stress    diagram 139 

with    parallel    chords 130 

with  non-parallel   chords 133 

Stresses   in 129 

Counter-bracing,  125,  127  ;  notation  128 
Couple,    defined,    6 ;    a    resultant, 

21 ;  moment  of 40 

Culman's    method 53 


Dead  load 69,  70 

on    bridges 214 

reactions,     roof 75 

reactions  and  stresses,  arch....  144 

stresses,  transverse  bent 152 

Deck   bridges .  210 

Deflection  in  beams 178 

Curve,    elastic 179 

one  end  fixed,  beam 190 

beam  two  ends  fixed 195 

diagram    184 

f  ormulffi,     beams 198 

problem,  plate  girder 184 

Determining   stresses 85 

Diagonals,   importance   of 126 

Main    127 

Diagram,   bending  moment 168 

Deflection,   184;   plate  girder...  184 

Force,   defined 6 

Moment  and  shear,  simple  beam  171 

for    shears 302 

Shear,  beams 168 

Space,   defined 6 

Stress    102 

stress  in  unsymmetrical  truss. .  116 


PAGE. 

for  web  stresses 306 

Wind   load 74 

Diagrams,  application  of 297 

Influence 267 

Different  polygons  for  same  forces     34 

Direction,  defined 5 

Distance  pole,   defined 20 

Duchemin's  formula  for  wind  pres- 
sure         73 

Dynamics,    definition 3 

E 

Eccentric   riveted   connection 164 

Economical   trusses 68 

Effective  reactions,   roof 78 

Elastic  curve 179 

End,  beam  with  one  fixed 189 

Leeward,  on  rollers 110 

Windward,    on   rollers 112 

Ends,  beam  with  two  fixed 195 

Fixed,  parallel  reactions 105 

Fixed,     horizontal,     components, 

reactions   equal 108 

of  columns,  conditions  of 150 

Engine   loads 267 

and    train    loads 285 

Equilibrant,   defined 6,     13 

Equilibrium,    defined 5 

of  concurrent  forces 12 

Conditions   for 26 

of  non-concurrent   forces 26 

Proolems    in 13,     27 

polygon     20 

of  system  of  forces 42 

Equivalence,    defined 6 

Equivalent  uniform  bridge   load.  . 

218,  219 

Exact  method,  moment  of  inertia.      61 

Expansion  in  bridges 213 

and    contraction 70 


Figure,  polygonal 125 

Fink    trusses 67 

Maximum  stresses  in 119 

Fixed  beam,  one  end 189 

columns    151 

columns,  stresses  in 157 

ends,    horizontal    components   of 

reactions   equal 108 

ends,   parallel,    reactions 105 

truss,    reactions 78 

Flange  stresses,  plate  girder 285 

Floor  beam,  maximum  reaction...  279 

system,    bridge 213 

Force,   definition 4 

diagram,    definition 6 

Moment    of 38 

polygon,    described 9 

triangle,     described 9 

Forces  at  a  joint 95 

Centroid  of  parallel 49 

Composition   of 6 

Concurrent,  resolution  of 11 

Concurrent    8 

Concurrent,    equilibrium    of..  12,  26 

Different  polygons  for  same.  ...  34 

Kinds  of,   defined 5 

Moment  of  system  of 42,  44 

Moments  of 85 

Moment  of  parallel 45 


INDEX. 


313 


PACK. 

Forces,    non-concurrent 16 

Non-concurrent,  resolution  of. . .  23 

Non-concurrent,  non-parallel. ...  16 

Forces,  one  side  of  section 97 

Parallel     22 

parallel,   moment  of  inertia  of.  53 

Resolution    of 6,  85 

Resultant  of  concurrent 9 

System    of 41 

Formulae  for  beams 198 

roof    weights 71 

wind     pressure 73,  74 

weight  of   bridges 214,  215 

Frame,    triangle 65 

Framed   structures 65 

Funicular    polygon 20 

Closing  of 23 

through   two  points,   35;   three.  36 


General    method    for    determining 

stresses 85 

Geometrical   areas 47 

Girder,  plate,  problem 184 

stresses   and   shears 285 

Grand  stand  truss,  stresses 160 

Graphic    determination,    radius    of 

gyration    58 

methods  for  beams 167 

methods  for  beam  deflections...    178 

transverse  bent 154 

moments : 43 

moments,   stresses   by 91 

resolution,   stresses   by 99 

statics,    defined 3 

Gravity,    center   of 47 

center   of,    irregular   area 50 

Gyration,  radius  of 57,     60 


Highway  bridges,  live  loads  for.  .  .  216 

Weights    of 214 

Hinged  arch 143 

Columns    151 

Columns,  stresses  in 155 

Horizontal    components 80 

of  reactions   equal 108 

Howe  bridge  truss 210 

roof    truss 81 

Button's    formula    for   wind   pres- 
sure     "3 


Inaccessible  points  of  intersection.     31 

Inclined  chords,  truss  with 280 

Inclined  chords 306 

surface,   wind  pressure   on <3 

Incomplete  framed  structure 66 

Influence  diagrams 267 

Inertia,   see  Moment  of  Inertia. 
Irregular     areas,     49;     center     of 

gravity    of 

Intersection,  inaccessible  points  of     31 
Interurban  bridges,  live  loads  for.   216 


PAGE. 

Jack  rafters 69 

Joint,    forces    at 95 

Position  of   loads   for  maximum 

moment    at 268,  271 

Joists,     bridge 213 

K 

Ketchum's       formula       for       roof 

weights    71 

Kinetics,   defined 3 

Knee-braces    213 

Known  lines  of  action 23 


Lateral  bracing 212 

systems,  wind  load  stresses  in.   255 
Leeward  end  of  truss  on  rollers. .   110 

rollers   81 

segment  of  arch,  wind  load.  .  .  .   148 

Line  of  action,  defined 5 

of  pressure  of  arch 35,     37 

Lines  of  action,  known 23 

Line,   load   and   moment  diagram.    295 

Live  loads  for  bridges 216 

Load,    dead 69 

Dead,       arch,       reactions       and 

stresses    144 

line   and   moment   diagram 295 

reactions,    Wind 78 

Snow,   on   roof 72 

Uniform,   on   bridge 217 

Uniform,  on  cantilever  beam...    169 
Uniform,  overhanging  beam....    177 

Uniform,  simple  beam 173 

Uniform,  beam  one  end  fixed...    193 

Wind 73 

Wind,  stresses 105 

Wind,  stress  on  arch 146,  148 

Wind,  stresses  in  lateral  systems  255 

Loaded   chords 298 

Loads  on  bridge 214 

Concentrated    168 

Concentrated,  simple  beam 171 

Concentrated,  one  fixed  end.  .  .  .   189 
Concentrated,  overhanging  beam  175 

Engine  and  train 267 

Maximum,   for   floor   beam  reac- 
tion        279 

Moving    199 

Position    of,    for    maximum    mo- 
ment    268,  271 

Position  of,  for  maximum  shear. 

275,  277 

Position  of,  for  maximum  stress 

in    web 280 

on   roofs 69 

on  lower  chord 103 

on    trusses 72 

on  upper  chord 99 

Wind,   on  bridges 219 

Long  compression  members 68 

Lower    chord 68 


M 


Magnitude,    definition 5 

Main  braces 127 


314 


INDEX. 


PAGE. 

Main   diagonals 127 

trusses,  bridge 211 

Maximum  and  minimum  stresses.  124 

stresses    114,  118 

chord  stresses,  Pratt  truss 291 

flange  stresses,  plate  girder....  285 

floor   beam    reactions 279 

moment,     concentrated     moving 

load 201 

moment    diagrams 297 

moment,  position  of  loads  for.. 

268,  271 

moment,  uniform  load 199 

shear    302 

shear,  plate  girder 285 

shear,  position  of  load  for. 275,  277 

shear,  two  loads 207 

shear,   uniform  load. 200 

stress,  inclined  chords 280 

stresses,   with  subordinate  brac- 
ing     308 

stress  in  web  members 280 

web  stresses,   Pratt  truss 291 

web    stresses 306 

Members  of  bridge  truss 211 

Web,    denned 68 

Merrinian's      formula      for       roof 

weight 71 

Methods    for    moment    of    inertia, 

Accurate,  61  ;  Approximate.  .  .  59 

Method  of  coefficients  for  stresses.  259 

Culmann's    53 

for   determining   stresses 85 

Graphic,  for  deflection 178 

Mohr's    55,  63 

Minimum  and  maximum  stresses.  .  124 

Moment  area 39 

(See  also  Bending  Moment.) 

beam  with  both  ends  fixed 195 

Bending    167 

Bending,  simple   beam 171 

Constant,  of  inertia 178 

of    couple 40 

diagram  and  load  line 295 

formula  for  beams 198 

of  inertia 52 

of   inertia    of   areas,    58 ;    about 

parallel    axis 63 

of  inertia,   table 63 

of  inertia,  variable 1M 

Maximum,    tables 297 

Maximum,    concentrated   moving 

load 201 

Maximum,  uniform  load 199 

of  parallel  forces 45 

Position  of  loads  for  maximum. 

268,  271 

of  resultant   39 

of  system  of  forces 42,  44 

Moments,    defined 38 

Algebraic,  stresses  by 86 

of    forces 85 

Graphic,      43 ;      stresses      by 

graphic    91 

Stresses   by,   in   bridge   trusses.  .  238 

Motion,  defined 4 

Moving   loads 199 


X 


Negative    moments. 


PAGE. 

Non-concurrent  forces 5,  16 

Equilibrium  of 26 

Resolution  of 23 

Non-coplanar  forces,  defined 5 

Non-parallel  chords,  counterbraced  141 

Non-concurrent  forces 16 

forces,  problem 29 

Notation,  Bow's 86 

Counterbi-acing  128 

described  7 


One  end  of  truss  on  rollers 80 

Overhanging     beam,     concentrated 

loads    175 

uniform  load 177 


Parabolic  bowstring  truss.  ..  .210,  247 
Parallel  axes,  moment  of  inertia.  . 

56,     64 

chords,    counterbraced 139 

chords,  stresses  in  trusses  with.    259 

forces    22 

forces,  centroid  of 49 

forces,   moment   of 45 

forces,  moment  of  inertia  of.  ...      53 

forces,   problem 27 

reactions 79 

reactions,  fixed  ends 105 

Parallelogram,  area  of 47 

Particle,   definition 4 

Pedestals    213 

Permanent  loads  on  trusses 72 

Petit  bridge  truss 210 

Pin    connections 213 

connected  trusses 68 

Pitch  of  roof,  defined 68 

Plate    girders 297 

Problem    184 

Stresses  and  shears 285 

Point,    defined,    4 ;    of   application, 

defined    5 

Position  of  loads  for   maximum 

moment    at 268,  271 

Points,  inaccessible  intersection. .  .      31 
Polygon,      through      two,      35 ; 

through    three 36 

Pole,    defined,    20 ;    pole    distance, 

defined    20 

Polygon,  clpsed .  . 11 

Closing   funicular ,      23 

Equilibrium    20 

Force,    described 9 

Funicular    20 

through        two        points,        35 ; 

through    three 36 

Polygonal  figure 125 

Polygons,    different    for    same    fig- 
ures          34 

Portals    212 

1'osition    of    loads    for    maximum 

floor  beam  reactions 279 

for    maximum    moment,    concen- 
trated moving  loads 203 

of  loads  for   maximum   moment. 
268,  271 


INDEX. 


315 


PAGE. 

Position    of    loads    for    maximum 

shear    : 275,  277 

of    loads    for    maximum    stress, 

\yeb    members 280 

Positive  moments 38 

Pratt  roof  truss 67 

truss,   stresses  in 228 

bridge    truss 210 

truss,  coefficients  for 264 

truss,    maximum    cord    and    web 

stresses   291 

Pressure  line  of  arch 35,  37 

Wind 73 

Problems,  general 32,  46,  75 

in    equilibrium 13,  27 

Problem,  Algebraic  moments 88 

Graphic  method  for   deflections.  181 

Graphic   moments 92 

Graphic    resolution 104 

Leeward  end  on  rollers 110 

Maximum  stresses  in  Fink  truss  119 

Plate   girder 184 

Stresses  by  algebraic  resolution.  96 

Stresses  in  cantilever  truss 114 

Truss  with  non-parallel   chords, 

counterbraced     133 

Truss  with  counterbraced  paral- 
lel chords 130 

Unsymmetrical   truss 116 

Wind  load  stresses -.105,  108 

Windward  end  on  rollers 112 

Purlins,  69  ;   weight  of 70 


Quadrangular    truss 67 

Quadrilateral,  area  of 48 


R 

Radius  of  gyration 57 

Table    63 

of  area 60 

Rafter,   jack 69 

Rafters,  weight  of 70 

Railroad  bridges,   live   loads  for.  .  216 

Weights  of 215 

Rays,    defined 21 

Reactions  for  arch,  dead  load.  .  .  .  144 

Effective,  roof 78 

fixed   ends    108 

by  graphic  resolution 99 

Horizontal  components,  equal...  108 

Maximum,  floor   beam 279 

Parallel 79 

Parallel,  fixed  ends 105 

for  roof  loads 75 

for  single  load,  arch 143 

for  transverse  bent 154 

Wind  loads,   roof 78 

Redundant  frame 66 

Relation    between    different    poly- 
gons for  same  forces 34 

Resolution  of  forces 6,  85 

of  concurrent  forces 11 

of  non-concurrent  forces 23 

Resolution,  stresses  by  graphic...  99 

Algebraic    95 

Rest,   definition 4 

Restrained  beams 187 

Resultant  of  parallel  forces 22 


PAGE. 

a  couple 21 

defined    6 

Moment  of 39 

of    non-parallel,    non-concurrent 

forces   16,  18,     21 

of  several  concurrent  forces.  ...        9 

of  two  concurrent  forces 8 

Rigid  body,  definition 3 

Rise,    defined 68 

Riveted  connections 213 

Eccentric    164 

trusses    68 

Rollers,  leeward  end  on 110 

under   truss 80 

Windward  end  on 112 

Roof    (see  also  Trusses). 

construction    ' 69 

Effective   reactions 78 

loads   69 

Reactions    75 

trusses,  65 ;  types,  67 ;   weights 

of    71 

truss    stresses 84 

Rotation    4,     38 


Saw  tooth  roof 67 

Section,  forces  on   side  of 97 

Moment  of  inertia  of,  table....  63 

Radius  of  gyration,  table 63 

Segment,  area  of 48 

Leeward,  of  arch 148 

Windward  of  arch  truss 146 

Sector,  area  of 48 

Shear,    defined 85 

beams    168 

beams,  with  both  ends  fixed 195 

diagrams     296 

diagrams,  simple  beam 171 

Maximum,  two  loads 207 

Maximum,    for    uniform    moving 

load    200 

Maximum,  position  of  loads  for. 

275,  277 

Shears     302 

plate  girder 285 

Stresses  by.  in  bridge  trusses. .  .  238 

Sheathing    69 

Shingles,   weight  of 70 

Signs   of  stresses 86 

Simple   beam 171 

construction,     beam     with     one 

fixed    end 191 

Slate,  weight  of 70 

Snow   load 72 

reactions,    roof 77 

stresses,  transverse  bent 152 

Space  diagram,  definition 6 

Span,    defined 67 

Special  construction  for  funicular 

polygons   34 

Stand,  grand,  truss  stresses 160 

Steel    (see  also  Corrugated  Steel). 

Weight    of 70 

Statics,    defined 3 

Straight    line    formula    for    wind 

pressure    73 

Strain,    defined 84 

Stress,    defined 84 

diagram    102 

diagram,  unsymmetrical  truss. .  116 


316 


INDEX. 


..GE. 

Stress  in   trestle   bent 162 

Wind,  on  arch 146 

Stresses   by  algebraic  moments.  .  .  86 

by  algebraic  resolution 95 

for  arch,  dead  load 144 

in  bridge  trusses   (see  names  of 
trusses). 

in  cantilever  trusses 144 

and   coefficients,    Warren    truss.  264 

in  columns 155 

in  counterbraced  trusses. ..  .129,  139 

in  flange,  plate  girder 285 

in  grand  stand  truss 160 

by   graphic   moments 91 

by   graphic    resolution 99 

Maximum    114,  118 

Maximum  and  minimum 124 

Maximum,  with  inclined  chords.  280 

by  moments  and  shears 238 

in  roof  trusses 84 

with   subordinate  bracing 308 

in  transverse  bent 150,  151 

in  trusses,  method  of  coefficients  259 

in  trusses  with  parallel  chords.  259 

in  unsymmetrical  trusses 114 

Web 306 

Wind   load 105 

Wind  load,  in  lateral  systems. .  .  255 

Strings,  defined 21 

Stringers,    bridge 213 

Structures,    framed 65 

Strut,    defined 68 

Sway    bracing 70,  213 

Subordinate    bracing 273,  308 

System  of  forces,  moment  by .  .  42,  44 

Moment  of  resultant 41 

Floor,  of  bridges 213 

of    parallel    forces,    moment    of 

inertia     53 


Table   of    moment    of   inertia   and 

radius  of  gyration 63 

Tar  and  gravel  roof,  weight  of .  .  .  70 

Tensile  stress,  sign  of 86 

Tension,    defined 84 

Three  hinged  arch.  .- 143 

non-parallel,     non-current     com- 
ponents    23 

Through   bridges 209 

Tie,   defined 68 

Ties  for  tracks 213 

Tiles,   weight   of 70 

Tin,  weight  of 70 

Top   of  column 151 

Tooth,  saw,  roof 67 

Train   loads 267,  285 

Transformation  of  moment  area..  39 

Transverse  bent,  stresses  in 150 

Translation,    defined 4 

Trestle  bent 162 

Triangle,   area  of 48 

frame,  shaped 65 

Force,    described 9 

as  a  truss 125 

Truss  (see  also  Bridges  and  Roofs 
and  name  of  each  truss). 

Arch    143 

Fink,  maximum  stresses  in 119 

fixed  both  supports 78 


.    PAGE. 
Grand    stand 160 

with    inclined    chords 280,306 

on    rolk     80,  112 

Roof,    s.      -ses 84 

with   suL      linate   bracing 308 

triangle    125 

Unsymmetrical,    stress    diagram 
for    116 

Trusses,  bridge,  types  of 210 

Cantilever     114 

Counterbraced,  stresses  in 139 

Economical     68 

Loads    on 72 

Roof    65 

Weight    of 71 

Stresses  in,  with  parallel  chords  259 
with    parallel    chords,    counter- 
braced    139 

Stresses  in  counterbraced 129 

Unsymmetrical     114 

Two    non-parallel,    non-concurrent 

components       24 

Types  of  roof   n-uhJ  s 67 

of  bridge  trusses 210 


!U 

Uniform  load,  beam  fixed  one  end  193 

on  bridge 217 

on  cantilever  beam 169 

overhanging  beam 177 

loads,  simple  beam 173 

Unloaded    chords 300 

Upper  chord 68 

Loads  on 99 

Unsymmetrical  trusses 114 

truss,  stress  diagram 116 


V 

Variable  moment  of  inertia......    184 


W 

Warren    bridge   truss 210 

Coefficients   and  stresses   in ....  264 

Stresses    in 306 

Web  members,  defined 68 

Maximum   stress   in 280 

Maximum  stress  in,  Pratt  truss.  291 

Stresses     306 

Weights   (see  article). 

of  highways  bridges 214 

of  railroad  bridges 215 

Wheel  loads  on  bridges 217 

Whipple  bridge   truss 210 

Wind   load 73 

on   bridges 219 

reactions 78 

stresses  in  arch  truss 146 

stresses  fixed  column 157 

stresses  hinged  column 155 

stresses  in  lateral  systems 255 

Leeward  segment  of  arch.  .  .  .  148 

on  roof 105 

pressure    73 

Windward  end  of  truss  on  rollers 

82,  112 

segment,  stress  on  arch 146 

Wooden  shingles,  weight  of 70 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN     INITIAL     FINE     OF     25     CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  5O  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


DEC  14  1932 


DCT  291940 


LD  21-50ni-8,-32 


HHAY 


I92Z 


YU 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


' 


•f/-  - 


r; 


, 


